Question

For the cell,
$\ominus\text{Zn}_{(s)} | \text{Zn}^{+2}(1\text{M}) || \text{Ag}^{+1}(1\text{M}) | \text{Ag}_{(s)}^\oplus$
If concentration of $\text{Zn}^{+2}$ decreases to $0.1\text{ M}$ at $298\text{ K}$, then emf of cell}

• A. increase by $0.0592\text{ V}$
• B. decrease by $0.0592\text{ V}$
• C. increase by $0.0296\text{ V}$
• D. decrease by $0.0296\text{ V}$

Hint:

Decreasing product concentration shifts equilibrium forward, increasing cell potential.

Answer 1

The Correct Option is C

Step 1: Concept Use the Nernst equation: $E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$.

Step 2: Meaning For the reaction $\text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag}$, $n=2$ and $Q = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2}$.

Step 3: Analysis Initially, $\log Q = \log(\frac{1}{1^2}) = 0$.
Finally, $\log Q' = \log(\frac{0.1}{1^2}) = -1$.
Change in emf = $-\frac{0.0592}{2} (-1) = +0.0296\text{ V}$.

Step 4: Conclusion The emf increases by $0.0296\text{ V}$. Final Answer: (C)