Q25 / 150 · MHT CET 2026 (20 April Shift 2) · Chemistry
Determine the electrode potential of $\text{Sn}^{2+}(0.01\text{M})\ |\ \text{Sn(s)}$ at $25^\circ\text{C}$ if $E^\circ_{\text{Sn}}$ is $-0.136$ V.
- A$0.186$ V
- B$-0.186$ V
- Correct answerC$-0.195$ V
- D$0.195$ V
Solution
The reduction half-reaction is $\text{Sn}^{2+} + 2e^{-} \rightarrow \text{Sn(s)}$.
Using the Nernst equation:
$E = E^{\circ} - \dfrac{0.0591}{n} \log \dfrac{1}{[\text{Sn}^{2+}]}$
Substituting the given values ($n = 2$, $[\text{Sn}^{2+}] = 0.01\text{ M}$, $E^{\circ} = -0.136\text{ V}$):
$E = -0.136 - \dfrac{0.0591}{2} \log \dfrac{1}{0.01}$
$E = -0.136 - 0.02955 \log (100)$
$E = -0.136 - 0.02955 \times 2$
$E = -0.136 - 0.0591$
$E = -0.1951\text{ V} \approx -0.195\text{ V}$
Answer: $-0.195$ V
Using the Nernst equation:
$E = E^{\circ} - \dfrac{0.0591}{n} \log \dfrac{1}{[\text{Sn}^{2+}]}$
Substituting the given values ($n = 2$, $[\text{Sn}^{2+}] = 0.01\text{ M}$, $E^{\circ} = -0.136\text{ V}$):
$E = -0.136 - \dfrac{0.0591}{2} \log \dfrac{1}{0.01}$
$E = -0.136 - 0.02955 \log (100)$
$E = -0.136 - 0.02955 \times 2$
$E = -0.136 - 0.0591$
$E = -0.1951\text{ V} \approx -0.195\text{ V}$
Answer: $-0.195$ V
Hence, the correct answer is option: 'C'.