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Chemistry Question on Electrochemistry

The e.m.f. of a Daniell cell at 298 K is E1. Zn/SO4 (0.01 M) || CuSO4 (1.0 M)/Cu. When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. is changed to E2. What is the relationship between E1 and E2 ?

A

E1 > E2

B

E1 < E2

C

E1 = E2

D

E2 = 0 ≠ E1

Answer

E1 > E2

Explanation

Solution

The Nernst equation for a cell with two half-reactions can be written as:

E=E°−RTnF ln([Cu2+][Zn2+])E=E°−\frac {RT}{nF }\ ln (\frac {[Cu^{2+}]}{[Zn^{2+}]})

In this case, when the concentration of ZnSO4 is 0.01 M and CuSO4 is 1.0 M, you have:

E1=E°−RTnFln(1.00.01)E_1=E°−\frac {RT}{nF} ln (\frac {1.0}{0.01})

E1=E°−RTnF(2ln(10))E_1=E°−\frac {RT}{nF }(2ln (10)) .......(1)

Now, when the concentration of ZnSO4 is 1.0 M and CuSO4 is 0.01 M, you have:

E2=E°−RTnFln(0.011.0)E_2=E°−\frac {RT}{nF} ln (\frac {0.01}{1.0})

E2=E°−RTnF(−2ln(10))E_2=E°−\frac {RT}{nF }(-2ln (10)) .........(2)

Now, E1−E2E_1−E_2 = E°−RTnF(2ln(10))E°−\frac {RT}{nF }(2ln (10)) - E°−RTnF(−2ln(10))E°−\frac {RT}{nF }(-2ln (10))

On simplifying,

E1−E2E_1−E_2 = −2RTnF(2ln(10))−\frac {2RT}{nF }(2ln (10))

Since 2RTnF\frac {2RT}{nF }is a positive constant, we can see that E1 - E2 is negative. Therefore, E1 is greater than E2

So, the correct relationship between E1 and E2 is option (A): E1 >E2