MHTCET 11 Apr 2026 Shift 2 Paper

Concept:The Nernst equation relates the electrode potential to the concentration of the electroactive species.Explanation:For the reduction half-reaction,$$\mathrm{Cu}^{2+} + 2e^{-} \rightarrow \mathrm{Cu}$$two electrons are transferred, hence the number of electrons involved is $n = 2$.At 298 K, the Nernst equation for this electrode is written as,$$E = E^{\circ} + \frac{0.0591}{n} \log [\mathrm{Cu}^{2+}]$$This equation shows how the electrode potential shifts when the ion concentration deviates from standard conditions.Now, substitute the given values,$E^{\circ} = 0.34 \, \mathrm{V}$ and $[\mathrm{Cu}^{2+}] = 0.1 \, \mathrm{M}$.$$E = 0.34 + \frac{0.0591}{2} \log (0.1)$$Since $\log(0.1) = -1$, the equation becomes,$$E = 0.34 + 0.02955 \times (-1)$$$$E = 0.34 - 0.02955$$$$E = 0.31045 \, \mathrm{V} \approx 0.31 \, \mathrm{V}$$Therefore, the electrode potential at a $\mathrm{Cu}^{2+}$ concentration of $0.1 \, \mathrm{M}$ is approximately $0.31 \, \mathrm{V}$.Answer:The correct option is C, i.e., $0.31 \, \mathrm{V}$.