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

Step 1: Understanding the Concept:
The relationship between cell potential (\(E_{\text{cell}}\)) under non-standard conditions and standard cell potential (\(E^\circ_{\text{cell}}\)) is given by the Nernst equation.
Step 2: Key Formula or Approach:
The Nernst equation at \(298\text{ K}\) is: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log_{10} Q \] where \(n\) is the number of moles of electrons transferred, and \(Q\) is the reaction quotient.
Step 3: Detailed Explanation:
From the balanced redox reaction: \(\text{Zn}_{(\text{s})} \longrightarrow \text{Zn}^{+2}_{(\text{aq})} + 2\text{e}^-\) (Oxidation)
\(2\text{Ag}^+_{(\text{aq})} + 2\text{e}^- \longrightarrow 2\text{Ag}_{(\text{s})}\) (Reduction)
The number of electrons transferred is \(n = 2\).
The reaction quotient \(Q\) for this reaction is: \[ Q = \frac{[\text{Zn}^{+2}]}{[\text{Ag}^+]^2} \] (Note: pure solids like Zn and Ag are not included in the Q expression).
The problem states that the cell potential is less than \(E^\circ_{\text{cell}}\) by \(0.0592\text{ V}\), which means: \[ E^\circ_{\text{cell}} - E_{\text{cell}} = 0.0592\text{ V} \] Substitute this into the rearranged Nernst equation: \[ E^\circ_{\text{cell}} - E_{\text{cell}} = \frac{0.0592}{2} \log_{10} Q \] \[ 0.0592 = \frac{0.0592}{2} \log_{10} Q \] Dividing both sides by \(0.0592\): \[ 1 = \frac{1}{2} \log_{10} Q \] \[ \log_{10} Q = 2 \] \[ Q = 10^2 = 100 \] Now we test the given options to find which gives \(Q = 100\):
(A) \(Q = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100\)
(B) \(Q = \frac{1}{(0.01)^2} = \frac{1}{0.0001} = 10000\)
(C) \(Q = \frac{0.1}{(1)^2} = 0.1\)
(D) \(Q = \frac{0.01}{(1)^2} = 0.01\)
Only option (A) yields \(Q = 100\).
Step 4: Final Answer:
The correct condition is \([\text{Zn}^{+2}] = 1\text{M}\) and \([\text{Ag}^+] = 0.1\text{M}\).