Question

Which from following statements is true regarding the cell emf at 298 K for \( \ominus Ni_{(s)} | Ni^{+2}(0.01M) || Ag^{+}(0.01M) | Ag_{(s)} \oplus \)?

• A. less than \( E^{\circ}_{cell} \) by 0.0592 V
• B. greater than \( E^{\circ}_{cell} \) by 0.0592 V
• C. less than \( E^{\circ}_{cell} \) by 0.0296 V
• D. greater than \( E^{\circ}_{cell} \) by 0.0296 V

Hint:

$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q$. Pay attention to the coefficients in the balanced equation.

Answer 1

The Correct Option is D

Step 1: Cell Equation
$Ni + 2Ag^+ \rightarrow Ni^{2+} + 2Ag$. Here $n=2$.
Step 2: Nernst Equation
$E_{cell} = E^\circ_{cell} - \frac{0.0592}{2} \log \frac{[Ni^{2+}]}{[Ag^+]^2}$.
Step 3: Substitution
$E_{cell} = E^\circ_{cell} - 0.0296 \log \frac{0.01}{(0.01)^2} = E^\circ_{cell} - 0.0296 \log(100)$. $E_{cell} = E^\circ_{cell} - 0.0296(2) = E^\circ_{cell} - 0.0592$ V. (Wait, let's recheck the calculation). $\frac{0.01}{0.0001} = 100$. $\log(100) = 2$. $E_{cell} = E^\circ_{cell} - 0.0592$. Actually, let's look at the options. If the question implies $Ni^{2+}$ is $0.01$ and $Ag^+$ is $0.1$, it changes. But with $0.01/0.0001$: $E = E^\circ - 0.0592$. This matches (A). Wait, if $Q<1$, then $E>E^\circ$. Here $Q = \frac{10^{-2}}{(10^{-2})^2} = 10^2 = 100$. $E_{cell} = E^\circ - 0.0592$. (Matches A). However, if $Ag^+$ was $0.1$: $Q = 10^{-2}/10^{-2} = 1$. Let's re-verify: $Q = \frac{0.01}{0.01^2} = 100$. $E_{cell} = E^\circ - 0.0296 \times 2 = E^\circ - 0.0592$.
Step 4: Conclusion
The value is less by 0.0592 V.
Final Answer:(A)