Dear students, welcome you all to VTU E-Shikshana program. In our previous class, I have given brief introduction on electrochemistry. We have discussed and understood...
the basic concepts such as free energy, what free energy is and what is E cell, E naught cell and we understood the working. principle of galvanic cells and we also understood the differences between the electrolytic and electrolytic cells and galvanic cells. Then we derived Nernst equation for single electrode potential.
In today's class we will be solving numerical problems making use of Nernst equation. In last class we. We derived the Nernst equation for single electrode potential.
So, the mathematical expression of Nernst equation is for single electrode potential is E is equal to E naught plus 0.0591 divided by N log to the base 10 of Mn plus. So, this equation is known as the Nernst equation for single electrode potential. So, making taking use of this equation we shall solve the numerical problems. So I think this also I have told.
So Nernst equation for cell is see E cell is equal to E0 cell. I have taken a simple equation electrochemical equation A moles of A plus B moles of B give C moles of C. and D moles of D. A moles of A combine with B moles of B and the products are C moles of C and D moles of D. So EMF of the cell is E0 cell minus 2.303 RT divided by NF log C rise to C.
So C rise to C the products will come in the numerator will be represented as the products will be represented as the numerator and the reactants will be represented as the denominator. So log C rise to C D rise to D divided by A rise to A multiplied by B rise to B. So, we shall solve the problems making use of this Nernst equation. See problem 1. Give the representation of a standard cadmium copper cell. Write the electrode and cell reactions and calculate the standard EMF of the cell.
If standard electrode potentials of cadmium and copper are are minus 0.4 volt and 0.34 volt respectively. So first we need to understand the problem given. See, give the representation of... standard cadmium copper cell.
First we need to represent the cell. The cell is made of these two electrodes cadmium and copper. Write the electrode and cell reactions.
We need to give the anodic and the cathodic reactions. Calculate the standard EMF of the cell. So we need to calculate the standard EMF of the cell.
If standard electrode potentials of cadmium and copper are minus 0.4 volt and plus 0.34 volt respectively. So they have given the standard electrode potentials of cadmium and copper. The standard SRP of cadmium is minus 0.4 volt and that of copper is plus 0.34 volt. So first we need to represent the cell.
So the cell is made of cadmium, the two electrodes, cadmium and copper. So we need to decide which one is anode. We need to find out which one is anode, which electrode is anode and which metal acts as The metal which has low SRP, see among these two cadmium and copper given, cadmium has an SRP of minus 0.4 volt and copper has a potential of standard electrode potential of plus 0.34 volt.
Therefore cadmium is anode and copper acts as cathode. So This is the cell representation. So by convention, anode should be written on the left and the metal which acts as cathode should be written on the right. Therefore, Cd slash Cd2+.
told in the in our previous class that this line this single vertical rain represents the phase difference between a metal and its ions. So, C D separated by C D 2 plus two vertical lines these two vertical lines represent the salt bridge cu2 plus slash cu so initially cu2 plus ions will be there they get reduced as copper so electrode this is how the electrode uh i'm sorry this cell is represented cd slash cd2 plus Since it is a standard cell 1 molar, 1 molar, electrode reactions we need to give at the anode. We should remember one thing that it is always oxidation that occurs at the anodic site. So, cadmium, since cadmium is anode, cadmium undergoes oxidation to give Cd2+, Cd2+, and electrons are released. Those electrons migrate into the cathodic area and the Cu2+, ions that are present in the cathodic container get reduced as copper.
So current cannot flow from one point to another point unless there is a potential difference between the two points. So this potential difference between the two electrodes which causes the flow of electrons from an anode. anode to cathode or the flow of current in the opposite direction is nothing but the emf of the cell right. So electrons are released into the anodic released by the anodic side and those electrons travel into the cathodic area.
into the cathodic container and the electrons are taken up by Cu2 plus ions so that copper gets deposited on the electrode cathode. So the net cell reaction is Cd plus Cu2 plus gives Cd2 plus plus Cu. So how to calculate the standard EMF?
So E0 cell, this should be E0, E0 cell is equal to E cell as we all know is E cathode minus E anode, E right electrode minus E left electrode is the EMF of the cell. E0 cell is E0 cathode minus E0 anode. E0 cathode, E0 cathode is copper. So E0 Cu minus E0 Cd. So these two values are given.
E0 of Cd is minus 0, I am sorry, E0 of, yes correct, E0 of Cd is minus 0.4 volt and E0 of Cu is plus 0.34 volt. Therefore 0.34 minus 0.4, so this is minus 7.4 volt. So, the standard emf of the given cadmium copper cell is minus 0.74 volt. One more problem we shall solve. Write the cell representation, cell reaction and calculate the standard EMF of nickel silver cell, NiAg cell.
If E0 of Ni and E0 of Ag are minus 0.25 volt. and plus 0.8 volt respectively. Write the cell representation. So we need to represent the cell.
Cell reaction, we need to give the anode as well as the cathode. reactions as well as the net reaction, the net cell reaction we are supposed to write and calculate the standard EMF, standard EMF E0 cell, standard EMF of NiAg cell if E0 of Ni and E0 of Ag are 0, minus 0.25 volt and 0.8 volt respectively. So cell So, as I told, as I have already told, anode should be on the left and cathode should be written on the right. So, while representing the cell, this is anode and this is cathode.
So, how to find out, that also I have, how to find out which electrode is, which electrode acts as anode and which electrode acts as cathode. So, it depends on their, SRP values, standard reduction potential values, the values are given E0 of Na is minus 0.25 volt and E0 of Ag is 0.8 volt. The metal with low SRP acts as this is anode and the metal with among these two, this has high SRP value.
So, this acts as cathode. Therefore, N Ni slash Ni2 plus since it is a standard cell 1 molar Ag plus slash Ag. So this is the representation of this given nickel Ag cell.
Then the anodic reaction. Anodic reaction is always oxidation. Therefore Ni undergoes oxidation.
It forms Ni2 plus and electrons are released. And electrons migrate into the cathodic chamber and they are, see, Ag plus combines with electrons so that Ag is formed. See, Ag plus, plus electron give Ag.
So, this is the reaction. So, the net cell reaction is this multiplied by 2. Therefore it becomes 2Ag plus plus 2 electron give 2Ag. So electrons, electrons these two get cancelled. So the net cell reaction is Ni plus 2Ag plus give Ni2 plus plus plus 2Ag.
So standard EMF again this should be E0 cell. Standard EMF of this cell is E0 cathode minus E0 anode. So cathode is silver and nickel is the anode.
0.8 minus of minus 0.25 because the SRP of nickel given is minus 0.25 volt 0.8 minus of minus 0.25 is 1.05 volt. So we need to carefully understand the given problem then we should start solving the problem. We shall solve one more problem making use of Nernst equation.
So for the cell Fe slash Fe2 plus of 0.01 molar. Ag plus of 0.1 molar slash Ag, write the cell reaction and calculate the MF of the cell at 298 Kelvin. If standard electrode potentials of iron Fe and Ag electrodes are minus 0.44 volt and the 0.8 volt respectively. For the cell Fe slash Fe2 plus slash Ag plus slash Ag, write the cell reaction. Calculate the MF of the cell at 298 Kelvin.
If standard electrode potential, the standard electrode potentials of iron and Ag are given and their values are minus 0.44 volt and 0.8 volt respectively. So E0 of... F e is equal to minus 0.44 volt E naught of Ag, right, silver, correct, is 0.8 volt.
So cell is already given. This is the given cell, Fe slash Fe2 plus slash Ag plus slash Ag. So first we need to write the cell reactions, anodic reactions, reaction.
the tucker at the anode and the reaction that takes place at the cathode. So Fe gives Fe2 plus plus 2 electrons. This is the reaction that takes place at the anode.
Anode reaction is always oxidation. So here Here iron is this is anode iron is anode and silver is this is cathode. So Fe undergoes oxidation to give Fe2 plus and electrons are released and at the cathode Ag plus plus electron forms they form Ag at multiplied by 2. So this is the net reaction the cell reaction is. is Fe plus 2Ag plus give Fe2 plus plus 2Ag.
So E0 cell we all know that E0 cell is E0 cathode right E0 cell is E0 cathode minus E0 anode. E0 cathode minus E0 anode that is what I have written here cathode is silver and iron is anode cathode minus anode therefore E0 of Ag minus E0 Fe. So 0.8 this is the potential of.
what silver given 0.8 minus 0.44 volt. So E0 cell is 1.24 volt. This is the standard EMF E0 cell. But what we need to find?
What we need to find out? Calculate the EMF of the cell. We need to find out the EMF of the cell.
E0 cell just now we have calculated. So by applying Nernst equation, equation for an electrochemical reaction AA plus BB. I have already told AA plus BB, CC, AA plus BB, CC plus DD. So E is equal to E0 cell.
minus 2.303 RT divided by NF, sorry RT divided by 2.303 minus 2.303 RT divided by NF log C to the power C, D to the power B. a to the power a b to the power b. So, by applying Nernst equation for a cell e is equal to e0 minus.
So, these values I have already substituted the values of r and f and even the value of n. number of moles of electrons 2 then T 298 Kelvin. So 0.0591 divided by N log C, this is C and this is D.
C rise to C, this is 1, 1 moles of Fe2 plus ions. C rise to C, this is C, Fe2 plus 1. D rise to D, this is D. and small d is 2. So Ag rise to 2 divided by A rise to A, 1 Fe.
Therefore Fe Ag plus. This is A, this is B. So B rise to B. B is Ag plus rise to B.
This is 2. So E is equal to E0 minus 0.0591 divided by N log, so these two, this Fe2 plus and the C. Fe, molar concentration of Fe and Ag is equal to 1 for pure solids. Since it is 1, I have substituted the values 1 for this and this.
So this will be E is equal to E0 minus 0.0591 divided by N log Fe2 plus divided by Ag plus rise to 2. So, E is equal to E naught minus 0.0591 divided by n. The value of n is 2 here log 0. These two values are given 0.1. The concentration of Ag plus is 0.1 molar whereas the concentration of Ap2 plus is 0.01 molar.
We shall substitute those two values here. So, log 0.01 divided by 0.1 is to the power 2. So, this becomes 1.24 minus 0.0591 divided by 2. This is 2. The value of n is 2. Log 1. Log 1 is 0. Therefore, it becomes 1.24 volts. So, one more problem we shall solve. These are all quite simple and interesting. See, calculate the MF of the following zinc silver cell at 23 degrees Celsius.
If the concentration of zinc sulfate and AgNO3 are 0.18 molar, 0.0301 molar respectively. Given that E0 of Zn2 plus slash Zn is equal to minus 0.75 volt and E0 of Ag plus slash Ag is equal to 0.8 volt. So. Please understand the problem given calculate the EMF of the following they have given the cell say it is the cell is zinc cell is made of zinc and silver and the potential of the of the cell should be calculated at 23 degree Celsius.
It is not 25 degree Celsius, it is 23 degree Celsius. If the concentrations of zinc sulfate and the concentration values of zinc sulfate and Dijon water are given, they are 0. 0.18 and 0.0301 molar respectively and they have given these two values E0 of Zn2 plus slash Zn E0 of this zinc standard electrode potential of zinc and Ag are given and they are minus 0.75 volt and plus 0.8 volt respectively. So since we need to determine now which we need to find out. find out which electrode is I mean which metal is anode and which metal acts as cathode. The metal I have already told couple of times the metal with low SRP value tends to act as anode.
So this is anode this acts as anode and this is this cathode. So that is what I have written here since E0 of zinc is lower than the E0 of silver, zinc electrode acts as anode and silver electrode acts as cathode. So cell representation Z Zn slash ZnSO4 anyways even the concentration values are given. So write the concentration values. The concentration of zinc sulphate is 0.18 molar and the concentration of silver nitrate is 0.0301 molar.
Write the values of the electrolytes given. Now we shall write the electrode reactions at the anode. So this is anode.
Zinc is anode. Therefore zinc undergoes oxidation. Very simple oxidation reaction.
Zinc gives Zn2+. It forms Zn2+. And electrons are released. And at the cathode consumption of electrons takes place.
It is. plus ions of silver nitrate get reduced as silver. So Ag plus plus electron give Ag. So this is the cell reaction Zn plus 2Ag plus. So these two electrons get cancelled. So Zn plus 2Ag plus gives Zn2 plus plus 2Ag.
And which formula we should make use of? This E cell, the Nernst equation for us. cell E is equal to E0 minus 2.303, E is equal to E0 minus 2.303 RT divided by 2.303 RT divided by NF log.
C, C. So this is C, this is A, this is B, C, D. C rise to C, A, A. A moles of A, B moles of B, C moles of C, D moles of D. Right? So C, C. C rise to C is Zn2 plus 1. A, G. D is A, G. Ag rise to this small d, 2, divided by Zn.
A rise to A, small a is 1, 1 mole of zinc, correct? Ag plus rise to 2. So E0 cell, E0 cell is equal to E0 of, see, E0 of... E0 cell is equal to E0 cathode minus E0 anode.
So cathode is, here the cathode is silver. Therefore it is 0.8. minus minus of minus 0.75.
So, 0.8 minus of 0.76 I have taken this should have been 0.75 minus 2.303 or gas constant value 8.3143 T temperature. It is again there is a mistake in this 293 this should have been 293 divided by n, n value is 2 and the value of f is 96500. So, log so these two values are 1 the concentration of the zinc and Ag1 so this becomes 0.18 divided by 0.301 square. So 1.8 plus 0.75 is 1.55 volt minus 0. So, after calculating this becomes 0.0674. So, 1.55 minus 0.0674.
So, EMF of the cell is 1.4926 volt. So, next problem. Calculate the half cell potential at 298 Kelvin.
For the reaction Zn gives Zn2 plus plus 2 electrons. Given Zn2 plus the concentration of Zn2 plus is 0.1 mol and E0 is minus 0.76 volt. So calculate the half cell potential.
So, electrode potential we need to find at 298 Kelvin and the reaction is given. So, it is the oxidation of zinc Zn gives Zn2 plus plus 2 electron. The concentration of Zn2 plus is given.
It is 0.1 mol and E0 of zinc is minus 0.76 volt. So, solution Zn2 plus 0.1 mol. E0 0.76 volt and the value of n is 2. This is n, right? So, the value of n is 2. So, applying Nernst equation for single electrode potential E is equal to E0 plus 0.0591 divided by n log mn plus.
This we have. already derived. So this is the expression or this is the mathematical form of Nernst equation for single electrode potential E is equal to E0 plus 0.0591 divided by N log Mn plus. So the value of the standard electrode potential is given the standard electrode potential of zinc is minus 0.76 volt plus 0.0591 divided by N.
The value of N is 2. And log, so the concentration of Zn2 plus is also given, they have given the concentration of Zn2 plus, so 0.1. After substituting these values, we will be getting the electrode potential, the half cell potential, the potential of zinc electrode is minus 0.7895 volt. One more problem we shall solve with the help of Nernst equation. Calculate the MF of the cell in which the reaction is Mg plus 2Ag plus give Mg2 plus plus 2Ag.
So they have given the reaction. When Mg2+, the concentration, molar concentration of Mg2+, is 0.13, whereas that of the concentration of Ag+, is 1 into 10 to the power of minus 4 moles. given that E0 of Ag plus slash Ag is 0.8 volt and E0 of Mg2 plus slash Mg is minus 2.37 volt. First properly we need to understand the given problem.
They have what? They are asking us to calculate the EMF of the cell. They have given the cell reaction.
Cell reaction is given. See, this is the cell reaction. Mg plus 2Ag plus give Mg2 plus plus 2Ag.
This is the net cell reaction. And they have given the concentration values of Mg2 plus and Ag plus. The concentration of Mg2 plus is 0. 0.13 molar. Whereas the concentration of Ag plus is 1 into 10 to the power of minus 4 mole. So since the standard electrode potential of magnesium.
See we need to find out now which one is anode. Which electrode is anode and which electrode is cathode. So E0 of Mgc is lower than that of the E0 of Ag. So magnesium acts as anode.
So first we need to. represent the cell Mg slash Mg2 plus two vertical lines Ag plus slash Ag. Here mention the concentration values Mg2 plus concentration 0.13 and Ag plus concentration here requires to be mentioned.
Cell reaction at the anode Mg gives Mg2 plus plus 2 electrons and at the cathode Ag plus combines with the electrons 2 Ag plus plus 2 electrons give 2 Ag. So the net cell reaction is Mg2 Mg plus 2Ag plus give Mg2 plus plus 2Ag. So according to Nernst equation, we have already derived a couple of problems, similar problems. E is equal to E0 minus 0.0591 divided by N log CcddAaBb. Therefore E is equal to E0, E0 cell, E0 cell is equal to E0 cathode minus E0 anode.
The cathode is Ag plus, right? So, Ag plus E0 of Ag plus slash Ag minus E0. The anode is Mg.
Mg slash, see, Mg, Mg2 plus. Therefore, it is Mg2 plus slash Mg. So these two values are given 0.8 and minus 2.37, 0.8 minus of minus 2.37 minus 0.0591 divided by n.
log 0.13 see these two become 1 the molar concentrations of pure solids the molar concentration is 1. Mg2 plus divided by Ag plus. Substitute these values 0.13. The given values are for Mg2 plus the concentration is 0.13 and for Ag plus it is 1 into 10 to the power of minus 4. So we will be getting the final answer.
Here it is. Mf of the cell is 2.96 volt. So, one more problem based on Nernst equation. Calculate the concentration of nickel chloride in the nickel electrode having a potential of 0.1692 volt at 25 degree Celsius given that E naught of Na2 plus slash Na is equal to minus 0.14 volt.
So, account According to Nernst equation E of Ni2 plus slash Ni is equal to E0 of E is equal to concentration of nickel chloride we need to determine. Now we shall continue with problem 6. Calculate the MF of the cell in which the reaction is Mg plus 2Ag plus give Mg 2 plus plus 2Ag. When the concentration of Mg 2 plus is 0.13 molar and the concentration of Ag plus is 1 into 10 raised to the power minus 4 molar. Given that E0 of Ag plus slash Ag is 0.8 volt and E0 of Mg 2 plus slash Mg is minus 2.37 volt. volt.
First understand the data given. They are asking us to determine the EMF of the cell. So, we need to determine the, we need to find out the EMF of the cell.
The cell reaction is given. Mg combines with Ag plus ions and the product are Mg2 plus and Ag. The molar concentration of Mg2 plus and Ag plus are given 0.13, 1 e 2 10 to the 10 raise to the power minus 4 and the potential SRP values of silver. and magnesium are given. They are 0.8 volt and minus 2.37 volt respectively.
Since E naught of Mg 2 plus slash Mg is less than the E naught of Ag plus slash Ag, magnesium is anode and silver is cathode. So this is how the cell is represented. Mg slash Mg 2 plus slash Ag plus slash Ag and the concentration values are given.
So, write the represent the values the concentration values write the concentration values here indicate the concentration values here 0.13 molar and 1 into 10 to the power of minus 4. Now, we shall give the cell reactions at the anode Mg gives Mg 2 plus plus 2 electrons oxidation occurs Mg gives Mg 2 plus plus 2 electrons and at the cathode Ag plus plus electron give Ag. So, 2 Ag plus plus 2 electron give 2 Ag. So the net reaction is Mg plus 2 Ag plus give Mg2 plus plus 2 Ag.
So according to Nernst equation E is equal to E0 minus 0.0591 divided by N log C rise to the power C D rise to the power D divided by A rise to the power A B rise to the power B. C rise to the power C. This is stoichiometry of the product C and this is the stoichiometry of the product D and A rise to the power A, B rise to the power B.
So E is equal to E naught of cathode minus C naught. anode, right, E0 cathode minus E0 anode minus 0.0591 divided by N log. So these two values, the more the for pure solids as we know molar concentrations are 1. So E0 of Ag plus slash Ag minus E0 of Mg2 plus slash Mg minus 0.0591 divided by N log Mg2 plus divided by the molar concentration of Ag plus.
So 0.8. minus of minus 2.37 minus 0.0591 divided by n the value of n is 2 log substitute these values and calculate you will be getting the emf of the cell as 2.96 volt so 2.96 is the emf of this cell with the concentrations of Mg2 plus and Ag plus 0.13 molar and 1 into 10 to the power minus 4 molar respectively. So, we shall solve one more numerical problem making use of Nernst equation. Calculate the concentration of nickel chloride in the nickel electrode having a potential of 0.16942 volt at 20. 25 degree Celsius given that E naught of n a 2 plus slash n a is equal to minus 0.14 volt.
So, they have given the value this value they have given it is minus 0.14 volt. We We need to determine the concentration of nickel chloride, the concentration of Ni2 plus we need to find out. So according to Nernst equation E is equal to E0 minus 0.0591 divided by N log Mn plus or this equation also you can take.
So since it is 25 degree Celsius you can take this one itself E is equal to E0 plus 0.0591 divided by N. Logger. m n plus so or even if we are taking this 2.303 r t it is ok it is fine no problem then we will have to substitute these values that is it so e naught of n i 2 plus slash n i is given minus 0.14 volt and even this what is this the potential of what is this then having a potential of nickel see that value is also given it is minus 0.16942 so minus 0.16942 16942 is equal to minus 0.14 which is the standard electrode potential of nickel. This I have taken to the numerator therefore plus 2.303 I have substituted all these.
values 8.3143 then into 96500 divided by n the value is n. So, after substituting and calculating we will be getting the concentration as 0.1011 molar. So, the concentration should be what in terms of it should be moles right.
So, last problem. What is the concentration of Na2 plus in the cell given below at 25 degree Celsius if the EMF is 0.601 volt? Ni, see this is the cell given.
The cell is made of nickel and copper, Ni slash Ni2 plus. This concentration we need to find out. Cu2 plus concentration is 0.75 molar slash Cu. Given E0 of Ni slash Ni2 plus is 0.25 volt and E0 of Cu2 plus slash Cu is equal to plus 0.34 volt. So, first understand the data given.
We need to determine the concentration of this Na2 plus ions. And they have given the E0 values of Cu2 plus, Cu2 plus the electrode copper, Cu2 plus slash Cu plus 0.34 volt and Ni slash Ni2 plus. See?
It is the value plus 0.25 volt is E0 of Ni slash Ni2+. It is the oxidation potential. Since we need to consider the reduction potential, this will be minus 0.25 volt. This is the oxidation potential 0.25 volt. So E0 of Ni slash Ni2+, is 0.25 volt.
Therefore, E0 of Ni2 plus slash Ni is minus 0.25 volt. So, carefully understand the data given. Then, start solving the problem. So, this is minus 0.25 volt. At the anode, Na gives Na 2 plus Na and nickel undergoes oxidation to give Na 2 plus plus 2 electrons and the reaction that takes place at the cathode is Cu 2 plus plus 2 electrons give Cu right.
So, the net reaction is Na plus Cu 2 plus give Na 2 plus plus Cu. So, according to Nernst equation we all very well know E cell is equal to E naught cell minus 2.303 RT divided by N f. log Cc, C rise to the power C, D rise to the power D, A rise to the power A, B rise to the power B.
So, substitute these values, all the values given. We need to determine the concentration of Ni2 plus after substituting all these values and see log Ni2 plus we will be getting. So, after calculating the calculate the anti-log of it then we will be getting 0.3187 moles.
So, this is the concentration of Na2+.