MHTCET 13 Apr 2026 Shift 2 Paper

Concept:At the anode, oxidation always takes place, which means loss of electrons.Explanation:Oxidation is defined as the loss of electrons.Therefore, the reaction possible at the anode must have electrons on the product side.Check each option for electron loss:Option A: $\mathrm{F}_{2}+2 e^{-} \rightarrow 2 \mathrm{~F}^{-}$ — electrons are gained, so it is reduction.Option B: $2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_{2}+2 e^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}$ — electrons are gained, so it is reduction.Option C: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+1 e^{-}$ — $\mathrm{Fe}^{2+}$ loses one electron to form $\mathrm{Fe}^{3+}$, so it is oxidation.Option D: $\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(s)$ — electrons are gained, so it is reduction.Thus, the only oxidation reaction is given by Option C.Answer:Option C: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+1 e^{-}$