Question

The emf of Daniell cell at $298 K$ is $E_{1}$ $Zn \left| ZnSO _{4}(0.01 M )\right|\left| CuSO _{4}(1.0 M )\right| Cu$ When the concentration of $ZnSO _{4}$ is $1.0 M$ and that of $CuSO _{4}$ is $0.01 M$, the emf changed to $E_{2}$. What is the relation between $E_{1}$ and $E_{2}$ ?

• A. $E_1 = E_2$
• B. $E_2 = 0 \ne E_2$
• C. $E_1 > E_2$
• D. $E_1 < E_2$

Answer 1

Answer: (C)

$E_1 > E_2$

Answer 2

In a Daniel cell,

At cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)
At anode: Zn(s) → Zn²⁺(aq) + 2e⁻

Ecell = E°cell - $\frac{0.0591}{n}$ log $\frac{anode}{cathode}$

Zn| ZnSO4 (0.01 M)

CuSO4 | Cu (1.0 M)

In the equation, anode = Zn2+ . Cathode = Cu2+

Substituting the value:

E1 = E° - $\frac{0.0591}{n}$log$\frac{0.01}{1.0}$

= E° - $\frac{0.0591}{2} log 10^{-2}$

E° + 0.0591 V

Now, for E2 : Zn (1.0 M) Cu (0.01)

E2 = E° - \frac{0.0591}{2} $$log \frac{1}{0.01}

E°- $\frac{0.0591}{2} log^2$

E° - 0.0591 V

E1 > E2