electrochemistry is the study of generation of electricity from the energy released during spontaneous chemical reactions and the use of electrical energy to carry out chemical reactions that do not proceed spontaneously the study of electrochemistry helps us in creating new eco-friendly technologies because the reactions carried out electrochemically are energy efficient and less polluting a large number of metals such as sodium and magnesium compounds like sodium hydroxide gases like chlorine and fluorine and many other chemical substances are produced by electrochemical methods batteries and fuel cells convert chemical energy into electrical energy and I used on a large scale in various instruments and devices in human beings the communication between the cells and the transmission of signals through the cells to the brain and vice versa have electrochemical origins an electrochemical cell is a device that is capable of either producing an electric current due to chemical action or off producing chemical action due to the passage of electricity there are two basic types of electrochemical cells voltaic or galvanic cells and electrolytic cells in voltaic cells the chemical energy of the spontaneous redox reaction is converted into electrical energy voltaic cells are also called galvanic cells the electrical energy produced by such batteries can be put to useful work powering our cell phones radios and other devices in electrolytic cells electrical energy is required to carry out a non spontaneous chemical reaction when you recharge your cell phone battery you are actually running an electrolytic cell we will be focusing on voltaic cells in this module before we discuss how voltaic cells work let's review the basics oxidation occurs when one or more electrons are lost as a result the oxidation number of species increases shown here is an oxidation half-reaction silver metal can lose one electron to form a silver ion reduction is defined as the gain of electrons that the oxidation number of a species decreases when reduction occurs for example when a sink ion gains two electrons zinc metal is produced we talked about half-reactions because they only described half of what is happening if one species loses electrons then another species must gain electrons oxidation cannot occur without reduction and reduction cannot occur without oxidation in an electrochemical cell both oxidation and reduction must occur you having troubled remembering the definitions for oxidation and reduction just think of an oil trick oxidation is loss reduction is gain which of these represent oxidation which represent reduction equation 1 is an oxidation half-reaction an electron is lost and the iron to iron becomes iron three equation two is reduction the aluminum ion gains three electrons to form aluminium metal and the oxidation state of aluminium goes from plus three to zero equation three is a reduction equation electrons are gained equation four is oxidation two electrons are lost when building a voltaic cell one often starts with a half cell a half cell or a redox couple consists of a metal electrode immersed in an aqueous solution containing the metal ions a gobble half-cell is illustrated here two half-cells must be connected to build a voltaic cell the electrodes are connected by an external wire through a voltmeter and a switch this however is not sufficient for the voltaic cell to run a salt bridge connects the solutions the salt bridge allows migration of ions between the half cells while preventing direct contact between the half cells in this diagram the salt bridge is a u-shaped tube containing an aqueous salt solution such as potassium nitrate in some Vaughn type cells a Buddhist frit is used or both electrodes may be immersed in the same electrolyte solution when the switch is closed the circuit is complete and electrons move through the external wire the migration of electrons through the external wire creates a current that can be used to run a motor or some other electrical device the electrode where oxidation occurs is called the anode and the electrode we're reduction occurs is called the cathode to help you remember this think of an ox and a red cat oxidation occurs at the anode reduction occurs at the cathode you let's consider what is happening at the electrode electrolyte interface metal ions from the solution have a tendency to deposit on the electrode trying to leave the electrode positively charged simultaneously metal atoms from the electrode have a tendency to go into the solution as cations leaving the electrode negatively charged these two processes are competing depending on which process has the stronger tendency a charge separation results difference develops between the electrode and the electrolyte which is called electrode potential when the concentrations of all the species involved in a half cell is unity the electrode potential is known as standard electrode potential as per our UPA C convention standard reduction potentials are now considered as standard electrode potentials as mentioned earlier the anode is the electrode where oxidation occurs the anode has a negative potential relative to the solution by convention the anode is considered the negative electrode in a voltaic cell the cathode is the electrode where reduction occurs the cathode has a positive potential relative to the solution by convention the cathode is considered the positive electrode in a voltaic cell the voltage cell we've been discussing is called the Daniel cell the net ionic equation the Daniell cell is given here zinc metal reacts with copper to ions to form zinc ions and copper metal since oxidation is a loss of electrons the oxidation half-reaction has zinc metal losing two electrons to form zinc ions reduction is a game of electrons so the reduction half-reaction has copper to ions gaining two electrons to form copper metal we know that oxidation occurs at the anode Zink must be the anode Copple must be the cathode because reduction occurs at the cathode electrons travel through the external wire from the anode to the cathode by convention when drawing voltaic cells the anode compartment is shown on the left and the cathode on the right generally voltaic cells are represented using a shorthand notation instead of a diagram showing the half cells in this notation a galvanic cell is represented by putting a single vertical line between the metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge using this notation a Daniels cell may be represented as shown here under this convention the EMF of the cell is positive and is given by the potential of the half cell on the right hand side minus the potential of the half cell on the left hand side that is the EMF of the cell is equal to e right - e left using this convention the EMF of a Daniels cell may be calculated as shown in an operating voltage cell electrons flow from the anode to the cathode the current flows from the positive electrode the cathode to the negative electrode the anode this current flow takes place when the circuit is closed because a potential difference exists between the two electrodes the potential difference between the two electrodes is called the cell potential or the cell electromotive force or EMF for shot cell voltage is measured in volts the potential of an individual half-cell cannot be measured directly we can only measure the difference in potential for two half-cells to simplify this reduction potentials are measured relative to an arbitrary standard called the standard hydrogen electrode you will learn more about the measurement of electrode potential in the next module you have learned in the previous module that the potential of an individual half cell cannot be measured you can only measure the difference between two half cell potentials which gives the EMF of the cell if the potential of a half cell is arbitrarily chosen as the standard the potential of the other half cell with respect to the arbitrary standard can be determined arbitrarily the standard hydrogen electrode is chosen as the standard reference electrode the standard hydrogen electrode consists of a platinum electrode coated with powdered platinum inside it last tube the platinum electrode is dipped into a one molar solution of hydrogen ions and pure hydrogen gas is bubbled in at one bar of pressure the standard hydrogen electrode is designed to carry out the half reaction shown here a hydrogen ion plus one electron produces half a mole of hydrogen gas by convention the potential of the standard hydrogen electrode is assumed as zero volts when all ion concentrations are one molar reduction potentials are called standard reduction potentials or standard electrode potentials the standard electrode potential of an unknown half cell is determined experimentally by combining it with the standard hydrogen electrode the EMF of the resulting galvanic cell is determined using a voltmeter as the potential of the standard hydrogen electrode is assumed as zero volts the measured EMF itself becomes the standard electrode potential of the given half cell for example the standard electrode potential of the zinc electrode is determined by combining it with the standard hydrogen electrode and then measuring the EMF the EMF of the cell is calculated as the potential of the right electrode - the potential of the left electrode the EMF of the cell is equal to zero point seven six volts does the standard electrode potential of zinc is equal to minus 0.76 volts the negative value of the standard electrode potential indicates that zinc can reduce the H+ ions similarly the standard electrode potential of copper is calculated as shown here the positive value of standard electrode potential indicates that cupric ions get reduced more easily than H+ ions in other words hydrogen ions cannot oxidize copper does Coppa does not dissolve in HCl on the other hand copper dissolves in nitric acid due to the oxidation of copper by nitrate ions you the values of standard electrode potentials of some electrodes are shown in the table here if the value of e naught is greater than zero then the reduced form of the species is more stable than hydrogen gas if the value of e naught is less than zero then hydrogen gas is more stable than the reduced form of the species as can be seen from the table Florine with the highest standard electrode potential value of plus 2.8 seven volts shows the highest tendency to get reduced to fluoride ions hence we can say that fluorine gas is the strongest oxidizing agent and fluoride ion is the weakest reducing agent Lithium Ion on the other hand with the lowest electrode potential value of minus three point zero five volts shows the least tendency to get reduced to lithium metal does lithium-ion is the weakest oxidizing agent and lithium metal is the most powerful reducing agent in an aqueous solution you can also see from the table that as the value of the standard potential increases from the bottom to the top of the table reactions show a greater tendency to undergo reduction the strongest oxidizing agent are found at the top why the strongest reducing agents are found at the bottom of the table as the value of standard potential decreases half-reactions have a greater tendency to oxidize the reactions in the table need to be flipped to write out the oxidation half-reaction and the sign for the electrode potential changes that is the standard oxidation potential is equal to minus standard reduction potential try to answer this question on your own before checking the answer which of these is the strongest oxidizing agent the nitrate ion elemental iodine all the dichromate ion answer the dichromate ion is the best oxidizing agent as shown in the table it has the most positive reduction potential sometimes inert electrodes such as platinum gold or graphite rods are used in voltaic cells inert electrodes provide a surface for electron transfer to occur but they don't participate in the reaction not electrodes are used then the electrodes do not change in size as the spontaneous reaction proceeds the potential difference between the two electrodes gives the cell potential for voltaic cells the cell potential is always positive to calculate the EMF of the cell add the potential for the cathode half reaction and the potential for the anode half reaction the electrode with the positive reduction potential acts as the cathode while the electrode with a negative reduction potential acts as the anode with respect to the standard hydrogen electrode you we develop our skills by completing a practice problem consider this situation a student constructs a voltaic cell by connecting a half cell containing silver wire immersed in 1m silver nitrate and a half cell made from aluminum wire immersed in 1m aluminium nitrate identify the anode and the cathode in the cell right the balance to net ionic equation that occurs calculate the EMF of the cell identify the direction of electron flow in the circuit identify the direction of ion movement through the salt bridge which electrode will get larger which electrode will get smaller there's a long to-do here so we work through the problem one piece at a time let's start by consulting our table of standard reduction potentials the silver half-reaction has a higher standard reduction potential than the aluminium half reaction this means that silver has a greater tendency to undergo reduction than aluminium since silver has the higher standard potential silver will undergo reduction aluminum has a greater tendency to oxidize so we will need to write an oxidation half-reaction for aluminium since reduction occurs at the cathode silver will be the cathode in this voltaic cell aluminum must be the a note since oxidation occurs at the anode now that we've identified the anode and the cathode writing the half-reactions is easy at the cathode a silver ion gains one electron to form silver metal the value for the standard reduction is plus zero point eight zero volts at the anode aluminium metal loses three electrons to form an aluminium ion compared to the standard reduction potential table we have flipped the reaction notice also that the sign of the electrode potential changes as well when we write the oxidation half-reaction we need to balance these half-reactions to show the conservation of change as well as Mars to do this we must make sure that the number of electrons lost equal the number of electrons gained in this problem we need to multiply the cathode half reaction by three however reduction potentials are intensive properties that the value of e not for the reduction half-reaction is unaffected now we are ready to find the overall balanced equation and calculate the cell potential adding up the two half-reactions gives the net equation three moles silver ions react with a mole of aluminium metal to form a mole of aluminium ions and three moles of silver metal to calculate the cell potential just add the cathode and anode potentials we get ESL is equal to plus two point four six volt you we have enough information now to draw a diagram of a voltaic cell electrons will flow through the external wire from the aluminium strip the anode to the cathode silver when the switch is closed an ions will flow from the salt bridge into the anode compartment [Music] and cations will flew into the cathode compartment as the cell runs the aluminium electrode will get smaller and the aluminum ion concentration will increase in the anode compartment the silver electrode will get larger why the silver ion concentration in the cathode compartment will decrease consider a voltaic cell such as the Daniell cell we studied earlier if an external opposite potential that is less than the cell potential is applied the reaction will continue in the forward direction if instead an external opposite potential is applied that is equal to the cell potential the reaction stops altogether there will be no current flow if the external opposite potential is larger than the cell potential the cell will now run as an electrolytic cell the diverse reaction will occur electrochemical cells have a variety of uses for example pH meters use electrochemical cells electrochemical cells can also be used to measure solubility product constant and other types of equilibrium constants their roles are useful in potentiometric titrations you standard electrode potentials are determined at 1m concentrations however cell potentials are concentration-dependent how can we determine the cell potential if the concentrations are not 1m Walter Nernst was a german physical chemist who won the nobel prize in 1920 one area of his research was the concentration dependence of galvanic cells the Nernst equation shown here allows us to calculate cell potential at non-standard conditions ESL is the cell potential at non-standard conditions Enoch is the cell potential at standard conditions our is the universal gas constant and T is the temperature in Kelvin also n is the number of electrons transferred in the equation and F is Faraday's constant 96,000 487 coulombs per mole q is the reaction quotient or the ratio of products to reactants in terms of their molar concentrations to calculate Q the products are put in the numerator and the reactants are in the denominator raise each species to a par to its coefficient in the balanced equation molarity czar used for concentrations as indicated by the square brackets remember that concentrations of pure solids and liquids are taken as unity by converting natural logarithm into base ten long and substituting the value of our F and T equal to 298 the equation can be reduces as shown here let's work through a typical problem involving the Nernst equation what is the cell voltage of a Daniell cell running with 0.1 molar zinc sulfate and 0.5 molar copper ii sulfate at 25 degrees celsius the balanced net ionic equation is given here here's a strategy for solving the problem first you need to find a cell at standard conditions using our table of standard electrode potentials then use the Nernst equation from the half-reactions we see that two electrons are transferred in the overall reaction therefore n equals two now we need to find inor cell by using the equation shown here Ino tar is the electrode standard potential of a right-handed or reductive electrode and II not l is the left-handed or oxidative electrode now work out the expression of queue action cogent since the concentrations of solids are taken as unity the reaction quotient Q becomes the concentration of zinc two ions over the concentration of copper two ions then we can substitute our expression for Q in the Nernst equation as shown here from here we can substitute our values for e not n and the ion concentrations into the Nernst equation the ratio 4q can be simplified as 0.1 divided by 0.5 equals 0.2 now it's a matter of simplification of the expression II cell is equal to one point one two volts pH meters also work on the principle of the Nernst equation yo is a typical problem you might encounter a hydrogen electrode is in contact with the solution with a pH of nine what is the potential of the electrode let's work through this problem together we'll start by listing what we know since we're told the pH is 9 the hydrogen ion concentration is 10 to the power of minus 9 molar the equation for the hydrogen electrode is h+ + 1 electron makes 1/2 hydrogen gas we assume that the pressure of hydrogen is held at one atmosphere from the equation we see that one electron is involved so N equals one by convention the standard potential is zero volts now we can substitute this information into the Nernst equation from the equation for the hydrogen electrode we see that Q equals H 2 over h plus by evaluating the equation we find that the electrode potential is minus 0.5 3:1 words let's start our discussion with the example of a Danielson as a Daniel cell starts functioning the concentration of copper two ions starts decreasing and that of zinc ions starts increasing it means that Q the reaction quotient in the Nernst equation is increasing this shows that the cell voltage will decrease at one stage the concentrations of both the ions become equal this indicates that the reaction will reach equilibrium and the voltmeter will read zero for this situation the Nernst equation can be written as shown here however at equilibrium equilibrium constant KC is equal to concentration of zinc two ions over the concentration of copper two ions substituting the values of R and F and of T as 298 Kelvin we get the equation for equilibrium constant KC as shown here consider a galvanic cell made from half cells using magnesium and nickel notice that the ion concentrations are all one molar what is the equilibrium constant of this cell try to solve this problem on your own before proceeding let's start with the half-reactions to find ISA since Neko has the greater reduction potential it is the cathode and magnesium is the anode by substituting the standard reduction potential values as shown in the equation here you can find that a cell is 2.1 one volt you can rearrange the equation that relates a cell to the equilibrium constant to solve for the log of Casey you should memorize this equation you can then substitute your information about this cell into the rearranged equation log of KC is equal to seventy one point five to find KC you can take the inverse log or the exponential function of both the sites this gives us a value of 3 point 1 6 into 10 to the past 71 for Casey an extremely large number it is also possible to derive an equation relating the Gibbs free energy and the cell potential the equation is shown here this is another equation that you should memorize note that standard cell potential is an intensive property but the Gibbs free energy is an extensive property the value of Delta G depends on the number of electrons transferred you you already know that the Gibbs free energy is related to the equilibrium constant by the expression Delta G is equal to minus rtlnk but these equations you can easily calculate the equilibrium constant and the Gibbs free energy for a given electrochemical cell let's consider a galvanic cell at 298 Kelvin this cell is made from half cells containing aluminium and sink where the ion concentrations are 1 molar from the table of standard reduction potentials you can easily find that the standard cell potential at these conditions is plus zero point nine zero volt we first need to look at the balanced equation since zinc has the higher electrode potential zinc undergoes reduction while aluminium undergoes oxidation notice that six electrons are transferred in the equation six electrons are lost to oxidize to aluminium atoms two aluminium ions therefore n is equal to six you know the Delta G is equal to negative NFE cell you can substitute these values in the equation we get the Gibbs free energy is equal to negative 521 kilojoules you can find Casey from a cell start with the equation relating ezel and Casey rearrange to find log of KC and substitute log of KC is equal to ninety one point five to five so we end up with a KC value of three point three five zero into 10 to the power 90 one with the galvanic cell the cell potential is always positive it means that the Gibbs free energy for a galvanic cell is negative which is consistent with a spontaneous process a spontaneous process favors products so we see large values of the equilibrium constant there are some key terms that you must first understand before proceeding to study the conductance of electricity the first key term is electrical resistance which has been given the symbol R it is the opposition to the passage of an electric current resistance is measured in ohms the SI base units for ohm are kilograms meter squared per second cubed into amperes squared the electrical resistance of any object is directly proportional to its length L and inversely proportional to its cross-sectional area E we need a proportionality constant to turn this proportionality into equality this proportionality constant is called resistivity or specific resistance and is represented by the Greek letter Rho the SI unit for resistivity is own meter quite often it's submultiple ohms centimeter is also used remember that one ohm meter is equal to 100 ohms centimeters or 1 ohm centimeter is equal to 0.01 ohm meters if the length of the conductor is one meter and the area of its cross section is one square meter then resistance R becomes equal to the resistivity ro dance resistivity is defined as the resistance of a substance when it is 1 meter long and has a cross-sectional area of one square meter conductance which has the symbol G is the inverse of resistance does we can write that G is equal to 1 upon R since R is equal to Rho into length upon cross-sectional area we can say that conductance is equal to area divided by Rho into length the unit for conductance is the Siemens which has been given the symbol s Seamans is equal to ohm inverse or MHO the reciprocal of resistivity is known as conductivity conductivity is represented by the lowercase Greek letter Kappa that we can write that Kappa is equal to 1 upon Rho we can substitute Coppa in the equation we wrote a little while ago for conductance a we can write that G conductance is equal to Kappa into cross-sectional area upon length if the length of the conductor is one meter and the area of its cross section is one square meter then conductance becomes equal to copper does conductivity is defined as the conductance of a material when it is one meter long and has a cross-sectional area of one square meter the SI unit for conductivity is Siemens per meter sometimes this is given as Siemens per centimeter note that one semen per centimeter is equal to 100 Siemens per meter the magnitude of conductivity varies greatly as shown in the table it depends primarily on the nature of the material although temperature and pressure can also affect conductivity depending on the magnitude of their conductivity materials are classified into conductors insulators and semiconductors you conductors are materials with large conductivities such as metals and metal alloys some non metals including graphite carbon black and some organic polymers also act as conductors insulators are materials with very low conductivity such as glass and ceramics semiconductors have conductivities between conductors and insulators these important electronic materials include silicon doped silicon and gallium arsenite apart from the three materials discussed there are other special types of conductors known as superconductors superconductors have zero resistivity or in finite conductivity the earliest materials identified as superconductors were metals and alloys at low temperatures below 15 Kelvin certain ceramics and mixed oxides show superconductivity at temperatures as high as 150 Kelvin now let's discuss about the conductivity in metals that is metallic conductance metallic conductance or electronic conductance is the electrical conductance through metals as the name indicates the conductance is due to the flow of electrons in the metal metallic conductance depends on the nature and structure of the metal the number of valence electrons per atom and the temperature of the sample let's move on to the conductance in electrolytic solutions you know that pure water exhibits conductivity although it is very low the concentration of hydrogen ions and hydroxyl ions in pure water is 10 to the power of minus 7 molar due to the presence of these ions in very low concentrations pure water has a very low conductivity when electrolytes are dissolved in water ions are released into the solution as the ions go into the solution its conductivity increases the conductance of electricity by ions present in a solution is called electrolytic conductance or ionic conductance the conductivity of aqueous solutions depends on several factors including the nature of the electrolyte the size of the resulting ions the salvation of the ions the concentration of the electrolyte and temperature directly measuring the conductivity of ionic solutions is problematic passing a direct current into an ionic solution changes the composition of the solution as electrochemical reactions may occur also unlike a solid conductor such as a metal wire we can't connect a solution to a Wheatstone bridge to measure the conductivity of an ionic solution we use a conductivity cell like the one shown in the illustration here a conductivity CIN contains two platinum electrodes coated with platinum black separated by a distance L with a cross-sectional area E an alternating current sauce is used as the source of power the column of the solution separating the two electrodes then has a resistance described by the equation R is equal to Rho into L upon E the cell constant G's star depends on the distance between the electrodes and their cross-sectional areas g star is equal to L divided by a the unit of cell constant is per centimeter if we know the length and the area we can calculate the cell constant directly however direct measurement of the length and area of cross section of the electrodes are both difficult and unreliable you a better approach is to determine the cell constant by measuring the resistance of the cell when it contains a solution of known conductivity for this purpose we use potassium chloride solutions whose conductivity is known accurately at various concentrations and temperatures the table here shows the conductivity of different concentrations of potassium chloride solution at 298 Kelvin thus a constant G Star is equal to resistance inter conductivity of the solution once we know the cell constant we can use it to measure the resistance or conductivity of any solution to measure the conductivity we can use a Wheatstone bridge this device contains two resistances labeled r3 and r4 and a variable resistance r1 the conductivity cell is the unknown resistance r2 the Wheatstone bridge is connected to an AC power supply and a suitable detector when the bridge is balanced no current flows through the detector at these conditions r2 is equal to r1 into r4 divided by r3 inexpensive conductivity meters are commercially available and can also be used to measure the conductance or resistance of the solution in a Cell once the sale constant and solution resistance are known we can calculate the conductivity of the solution by dividing the cell constant by the resistance the conductivity of different electrolytes in the same solvent and at a given temperature varies this has become some variations in the charges and sizes of the resulting ions the ion concentrations and the ease of ion movement in the presence of a potential gradient therefore we introduce the concept of molar conductivity which uses a Greek letter lambda with a subscript M as its symbol molar conductivity is related to solution conductivity by the equation lambda M is equal to Kappa or conductivity divided by concentration of the solution if concentration is expressed in moles per cubic meter and conductivity in Siemens per meter then the unit for molar conductivity can be written as Siemens meter square per more to use molarity which is a far more common concentration unit for chemists we need to convert the unit's resulting in the second equation given here let's work through a typical problem together a conductivity cell is filled with 0.1 m KCl the resistance of the KCl solution it's 200 ohm the same conductivity cell is filled with 0.03 m NE CL and the cell resistance is 500 ohm what is the conductivity and molar conductivity of 0.03 M NaCl the conductivity of 0.1 m kc n solution is one point two nine Siemens per meter note that there are multiple paths to the problem here's our strategy first we need to find the sale constant g-star can do this by using the information given about the KCl solution we know that G star is equal to conductivity multiplied by resistance given that the conductivity of 0.1 molar potassium chloride solution is 1.29 Siemens per meter remember that Siemens is equal to inverse ohms the resistance of the solution is measured to be 200 ohms therefore thus a constant is the product of 1.29 times 200 the unit of ohms cancels giving a cell constant of 258 inverse meters or 2.58 inverse centimeters with the cell constant we can now find the conductivity of the 0.03 molar sodium chloride solution we know that conductivity is equal to the cell constant divided by the resistance of the solution which has been given as 500 ohms so we calculate that the conductivity of the 0.03 molar nacl solution is zero point five one six Siemens per meter we can use the equation shown here to find the molar conductivity from the conductivity and the molarity of the sodium chloride solution by substituting and evaluating we determine the molar conductivity to be one point seven two into ten to the path all Siemens meters where'd burn more you the conductivity and molar conductivity of an electrolyte change with the concentration of the electrolyte as concentration decreases conductivity decreases for both strong and weak electrolytes this is because there are fewer ions per unit volume to carry current as the solution is diluted however molar conductivity increases as concentration decreases the variation is different for strong electrolytes and weak electrolytes before we discussed the concentration dependence of molar conductivity we must first understand the idea of limiting molar conductivity this is the conductivity of a solution as its electrolyte concentration approaches zero limiting molar conductivity is given the symbol lambda M naught the molar conductivity of strong electrolytes is weakly dependent on concentration if we plot a graph of molar conductivity against the square root of concentration we see that the y-intercept is the limiting molar concentration the slope of the resulting line is the Coolridge coefficient which depends on the specific salt the cold rash coefficient depends on the cation and anion charges of the ions formed by the electrolyte you can classify electrolytes by the charges formed for example NaCl is considered to be a 1:1 electrolyte since the sodium ion is +1 and a chloride ion is minus 1 using similar reasoning you can classify calcium chloride as a 2-1 electrolyte magnesium sulfate is a to to electrolyte all electrolytes of a particular type have the same cold ROG coefficient called rosh carried out studies on limiting molar conductivity for a variety of electrolytes from his work he did used the law of independent migration of ions or coal Russia's law this law states that limiting molar conductivity can be represented as the sum of the individual contribution of the anions and cations of the electrolyte for assault of general form a why be sent at infinite dilution limiting molar conductivity is equal to Y times the limiting molar conductivity of cation a plus Z times the limiting molar conductivity of anion be try to work through this problem on your own before checking the answer calculate the limiting molar conductivity for sodium sulfate from the data given here sodium sulfate contains two sodium ions and one sulfate ion from cold rashes law the limiting molar conductivity is equal to two times the limiting molar conductivity of sodium ions plus the limiting molar conductivity of the sulphate ions substituting the values from the table we get the limiting molar conductivity of sodium sulfate as two hundred and sixty point two Siemens centimeter square per mole you the molar conductivity of a weak electrolyte depends greatly on the concentration of the solution this is because weak electrolytes have a low degree of dissociation as concentration increases since the relationship isn't linear we can't determine the limiting molar conductivity by extrapolating back to zero concentration at infinite dilution the electrolyte dissociates completely but the conductivity is so low that it can't be measured directly instead we can use cool raschi's law to find the limiting molar concentration we can approximate the degree of dissociation alpha for a weak electrolyte as the ratio of molar conductivity to limiting molar conductivity if the molar conductivity and the limiting molar conductivity known then the dissociation constants for weak electrolytes at a given concentration can be found typical problem conductivity of a weak acid in solution with a concentration of zero point zero zero two zero three four molar is 5 point 6 7 into 10 to the power minus 5 Siemens per centimeter calculate its dissociation constant if the limiting molar conductivity is 452 Siemens centimeter square per mole try to solve this problem on your own before checking the solution first find the molar conductivity using the equation you learned earlier on substituting the values we get the molar conductivity as twenty seven point eight eight Siemens centimeter square per mole now you can find the degree of dissociation for this weak acid since you know that alpha the degree of dissociation is the ratio of molar conductivity and limiting molar conductivity by substituting the values and simplifying we get the degree of dissociation as zero point zero six one six eight with the value of the degree of dissociation in hand we can use the equation relating ka and Anza substituting the values we get ka is equal to 8 point 2 5 into 10 to the power minus 6 an electrolytic cell is an arrangement in which an external source of voltage dreams about a chemical change since electrical energy is put in the reaction is non-spontaneous when you recharge a battery you're actually running them as electrolytic cells let's look at a simple electrolytic cell as shown in the diagram here it consists of two complex roads immersed in an aqueous copper sulphate solution the electrodes are connected to a DC power supply when the power supply is turned on reduction occurs at the cathode which is the electrode connector to the negative lead off the pass supply oxidation occurs at the anode which is connected to the positive lend of the past supply the reduction half reaction at the cathode has copper to ions gaining two electrons to form metallic copper at the anode copper metal undergoes oxidation by losing two electrons to form copper 2 ions students often find it difficult to remember the difference between galvanic and electrolytic cells in galvanic cells the anode is considered to be the negative electrode and reactions are spontaneous the cell potential is always positive and the change in Gibbs free energy is negative in electrolytic cells the cathode is considered to be the negative electrode and the reactions are non-spontaneous the cell potential is negative and the change in Gibbs free energy is positive remember in any cell oxidation occurs at the anode and reduction occurs at the cathode for example an impure sample of copper can be used as the anode as sailed runs pure copper is deposited on the cathode less reactive metals such as silver and gold and non reactive impurities fall to the bottom as a notes lunch similar processes are used to produce sodium magnesium aluminum and other metals on large scales michael faraday was the first scientist to quantitatively study electrolysis Faraday's work can be summarized in his two loans of electrolysis Farra DS first law states that the extent of chemical reaction that occurs at any electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte the electrolyte may be molten or in an aqueous solution in other words the mass of an element that deposits at an electrode during electrolysis is proportional to Q the total amount of electricity passed through the cell Faraday's second law states that the amounts of different substances produced by the same amount of electricity passing through the electrolytic solution are proportional to their equivalent masses equivalent mass can be calculated by dividing the atomic mass of a metal by the number of electrons needed to reduce a cation the amount of charge q passed through the cell can be determined from the product of current and time to determine the charge in coulombs multiply the current in amperes by the number of seconds for which it was applied the stoichiometry of the half reaction needs to be considered in quantitative electrolysis the half-reaction relates the number of electrons to the metal deposited for example in the reduction of silver ions one mole of electrons is needed for each mole of silver ions that reduce you know that the charge on one electron is equal to one point six zero to one multiplied by 10 raised to the power of minus 19 coulomb hence the charge on one mole of electron is equal to the product of the charge on an electron and the Avogadro number this can be written as the product of one point six zero to one multiplied by 10 raised to the power of minus 19 and 6.02 x 10 raised to the power of 23 therefore the charge on one mole of electrons is equal to ninety six thousand 487 coulombs per mole this quantity of electricity is called faraday which is symbolized by capital f sometimes this is rounded off to 96,500 coulombs per mole of electrons now use all this information you just learned to solve a problem a 2.50 ampere current is passed through molten magnesium chloride for 15 points zero minutes what is the mass of magnesium deposited at the cathode let's work through this problem together first we find the charge that passed through the cell you know that charge in coulombs is equal to the current in amperes multiplied by the time in seconds we need to convert 15.0 minutes into seconds 15 minutes into 60 seconds per minute is equal to 900 seconds the current is 2.50 amperes the charge that passed through the cell is then 900 seconds times 2.50 amperes which is equal to 2250 coulombs from half-reaction we see that two moles of electrons are needed for each mole of magnesium atoms that form therefore we require to faraday's or two multiplied by ninety six thousand four hundred and eighty seven coulombs to deposit one mole or 24 grams of magnesium therefore for 2250 coulombs the mass of magnesium deposited is equal to 24 grams per mole multiplied by 2250 coulombs divided by 2 multiplied by 96 thousand 487 does 0.279 eight grams of magnesium is deposited at the cathode the products of electrolysis depend on the nature of the electrolyte being electrolyzed and the type of electrodes used inert electrodes do not participate in the chemical reaction inert electrodes at as a sous on a sink for the electrons providing a surface for the electron transfer Platinum Gold or graphite electrodes are commonly used as inert electrodes on the other hand active electrodes do participate in the chemical reaction active electrodes are used in the electrolytic refining of metals which we discussed earlier does we can conclude that the products of electrolysis may be different for reactive and inert electrodes the products of electrolysis reactions depend on the oxidizing and reducing spacings present in the electrolytic cell and their standard electrode potentials some electrochemical processes are kinetically very slow at lower voltages these processes don't seem to occur at all in order to make these processes occur extra potential called over potential must be applied for example the electrolytic decomposition of molten sodium chloride will produce sodium metal and chlorine gas in this case only one cation that is the sodium ion and one anion that is the chloride ion are produced sodium ions are reduced to sodium metal at the cathode why chloride ions are oxidized to chlorine gas at the anode the electrolytic decomposition of aqueous sodium chloride solution on the other hand Fred uses sodium hydroxide guess and hydrogen gas in this case besides sodium and chloride ions hydrogen ions and hydroxide ions along with water molecules are present let's look at this process in detail at the cathode either sodium ions or hydrogen ions may be reduced as the hydrogen ion has a higher standard electrode potential than sodium it gets reduced hydrogen ions are produced by the dissociation of water hence the net reaction at the cathode meetly written as the sum of these two reactions does we have the reduction of water as per the equation shown here at the anode - there are two substances that can be oxidized chloride ions and water molecules at first glance it would seem easier for water to get oxidized dan chloride ions since the electrode potential of water is lower than that of chloride ions however the oxidation of water is a kinetically slow process hence an extra potential called over potential is needed for the production of oxygen this explains why chlorine but not oxygen is produced during the electrolysis of aqueous sodium chloride the net reactions of electrolysis of an aqueous solution of sodium chloride are as shown here it is more challenging to predict the products of electrolysis reactions in aqueous solutions due to over voltages you can't rely solely on standard production potentials production will occur at the cathode but there are two possibilities for the reduction half-reaction first ask yourself if the cation is a transition metal or a representative metal cation transition metal cations will undergo reduction since they have favorable reduction potentials representative metal cations will be too difficult to reduce instead water will be reduced at the cathode forming hydrogen gas predicting the oxidation half-reaction x' is more complicated it depends on the relative ease with which the different species present in the solution are oxidized halide ions will undergo oxidation to form the diatomic elements in preference to water however chloride ions undergo oxidation in preference to water due to over voltage of oxygen on the other hand opsal anions are too difficult to oxidize so water is oxidized to form oxygen gas the ability of a cation to be reduced all in an ion to be oxidized that only depends on the standard reduction potentials but also on their concentrations hence to account for the effects of concentration the standard electrode potentials are replaced by the electrode potentials given by the Nernst equation in dilute solutions of sulfuric acid water will be oxidized to release oxygen gas however in concentrated solutions of sulfuric acid the formation of the peroxy dye sulfate ion is favorite let's test our knowledge by solving some practice problems try to answer this question on your own before proceeding predict the products and write a balanced equation if an aqueous solution of silver nitrate is electrolyzed assume that the electrodes are inert let's start with the reduction process at the cathode silver a transition metal has a favorable reduction potential so silver ions will be reduced to form silver metal the half-reaction is written out since the electrode potential value of water is less than that of the nitrate ion water will undergo oxidation to form oxygen gas as shown in the half reaction now we can sum the two half-reactions we need to make sure that the number of electrons gained is equal to the number of electrons lost and then and the half-reactions to write the overall equation here's another problem try to answer it on your own first a dilute solution of sodium sulfate is electrolyzed predict the products of the electrolysis assuming that the electrodes are inert we use the same basic strategy as the previous problem at the cathode it is easier to reduce water than the sodium ion as it has a higher standard reduction potential value than sodium so water will undergo reduction to form hydrogen gas at the anode it is easier to oxidize water than the sulfate ion since the standard reduction potential for water is less than that of sulfate ions water will oxidize to produce oxygen gas we need to make sure that the number of electrons gained is equal to the number of electrons lost by summing the two half-reactions we can write an overall equation at h plus plus o h minus make h2o you if you have ever observed rusted iron tarnished silver or the green patina that develops on copper or bronze then you are familiar with corrosion corrosion is the term used to describe the process of the surface of metal objects getting covered by oxides or other salts of the metal in corrosion the metal is oxidized by losing electrons to oxygen forming oxides corrosion is essentially an electrochemical process let's look at the corrosion of iron which is commonly known as rusting in detail iron corrodes in the presence of water and air the spot where the corrosion begins can be considered the anode here iein loses two electrons to form the iron ii ion the oxidation half-reaction has an oxidation potential of minus 0.4 for volt the electrons dance released then move to another spot on the metal which then behaves as the cathode at this spot oxygen is reduced in the presence of a hydrogen ion to form water the hydrogen ions are believed to be available either from carbonic acid that is formed when carbon dioxide dissolves in water or due to the dissolution of other acidic oxides from the atmosphere the reduction potential that this process is 1.2 3 volts in the overall equation iron is oxidized to iron to ions while oxygen is reduced to form water the cell potential is 1.67 volts consistent with the spontaneous reaction the ferrous ions formed are been further oxidized by atmospheric oxygen to ferric ions in the form of hydrated iron 3 oxide commonly known as rust it is very important to prevent corrosion because it damages electrical tones bridges buildings which can collapse as a result and ships which can develop a breach corrosion can also force major industrial plants such as power plants or chemical processing plants to be shut down hence corrosion not only results in loss of money but also in dangers Public Safety how can corrosion be prevented one way is to cover the surface with paint all some chemicals like bisphenol this creates a physical barrier that protects the metal a second alternative is to cover the surface with other metals such as Tim all sink like paint this prevents oxygen and water from reaching the metal surface we can also use electrochemistry to prevent corrosion in cathodic protection the metal is ported with another metal that is more easily oxidized in cathodic protection of iron iron becomes the cathode and the metal coating becomes the sacrificial anode cathodic protection can be used to protect steel pipes from corrosion magnesium oxidizes more readily than iron as you know from the table of standard reduction potentials in this diagram a magnesium rod is buried and soldered to a steel pipe magnesium acts as the anode and the pipe acts as the cathode the damp soil acts as an electrolyte the Magnusson rod can be dug up and replaced this is much faster and less expensive than replacing the pipe it sells try to answer this question on your own before proceeding based on the table of standard reduction potentials which of these metals could provide cathodic protection to iron aluminium copper nickel or sink answer aluminum and zinc could be used to protect iron from corrosion as shown in the table their reduction potentials are more negative than that you're fine this means that they are more readily oxidized than iron copper and nickel reduce more readily than iron making them unsuitable for cathodic protection [Music] [Applause] [Music] [Applause] [Music] [Applause] 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