Question

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 1

The Correct Option is A

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

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

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

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

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

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

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

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

Now, \(E_1−E_2\) = \(E°−\frac {RT}{nF }(2ln (10))\) - \(E°−\frac {RT}{nF }(-2ln (10))\)

On simplifying,

\(E_1−E_2\) = \(−\frac {2RT}{nF }(2ln (10))\)

Since \(\frac {2RT}{nF }\)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