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:

Apply the Nernst equation and find the reaction quotient for which the cell potential drops by 0.0592 V.

Answer 1

The Correct Option is A

Step 1: Write the expression
$E=E^{\circ}-0.0296\log Q$, where $Q=\frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2}$.

Step 2: Required drop
A drop of 0.0592 V means $0.0296\log Q=0.0592$, so $\log Q=2$.

Step 3: Match
With $[\text{Zn}^{2+}]=1\text{ M}$, we need $[\text{Ag}^{+}]^2=0.01$, so $[\text{Ag}^{+}]=0.1\text{ M}$, the first option.

Final Answer:
The correct concentrations are 1 M $\text{Zn}^{2+}$ and 0.1 M $\text{Ag}^{+}$. \[ \boxed{\text{(A) }[\text{Zn}^{2+}]=1\text{ M and }[\text{Ag}^{+}]=0.1\text{ M}} \]