Question

For the cell reaction,
\(\text{Zn}_{(\text{s})} + 2\text{Ag}^+_{(\text{aq})} \longrightarrow \text{Zn}^{+2}_{(\text{aq})} + 2\text{Ag}_{(\text{s})}\)
Cell potential is less than \(\text{E}^\circ_{\text{cell}}\) by \(0.0592\text{ V}\) at \(298\text{ K}\) when

• A. \([\text{Zn}^{+2}] = 1\text{M}\) and \([\text{Ag}^+] = 0.1\text{M}\)
• B. \([\text{Zn}^{+2}] = 1\text{M}\) and \([\text{Ag}^+] = 0.01\text{M}\)
• C. \([\text{Zn}^{+2}] = 0.1\text{M}\) and \([\text{Ag}^+] = 1\text{M}\)
• D. \([\text{Zn}^{+2}] = 0.01\text{M}\) and \([\text{Ag}^+] = 1\text{M}\)

Hint:

If decrease in EMF is \(0.0592\) at 298K, then \(\log Q = n\) $\Rightarrow$ \(Q = 10^n\).

Answer 1

The Correct Option is B

Concept:
Nernst equation at \(298\,K\): \[ E = E^\circ - \frac{0.0592}{n}\log Q \] where \(Q\) is reaction quotient. Step 1: Write expression for \(Q\). \[ \text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag} \] \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \]
Step 2: Use given condition. Given: \[ E^\circ - E = 0.0592 \] So, \[ 0.0592 = \frac{0.0592}{n}\log Q \] Here \(n = 2\) (2 electrons transferred), hence: \[ 0.0592 = \frac{0.0592}{2}\log Q \]
Step 3: Solve for \(Q\). \[ \log Q = 2 \Rightarrow Q = 100 \]
Step 4: Check options. For option (B): \[ Q = \frac{1}{(0.01)^2} = \frac{1}{0.0001} = 100 \]
Step 5: Conclusion. Hence, correct condition is: \[ [\text{Zn}^{2+}] = 1\,M,\quad [\text{Ag}^+] = 0.01\,M \]