JEE Advanced 2023 — Paper 2 — Question 31

Oxidation half: $\mathrm{H_2S \xrightarrow{} S + 2H^+ + 2e^-}$

Reduction half (acidic): $\mathrm{MnO_4^- + 8H^+ + 5e^- \xrightarrow{} Mn^{2+} + 4H_2O}$

Equalizing electrons (LCM = 10), thus we have

Oxidation ×5: $5\,\mathrm{H_2S \xrightarrow{} 5S + 10H^+ + 10e^-}$

Reduction ×2: $2\,\mathrm{MnO_4^- + 16H^+ + 10e^- \xrightarrow{} 2Mn^{2+} + 8H_2O}$

On adding, we have $5\,\mathrm{H_2S + 2MnO_4^- + 6H^+ \xrightarrow{} 5S + 2Mn^{2+} + 8H_2O}$

Scaling to $8\,\mathrm{H_2S}$, thus factor = $\dfrac{8}{5}$

Water formed $= 8 \times \dfrac{8}{5} = 12.8 \Rightarrow 12.8\,\mathrm{mol}$

Electrons transferred $= 10 \times \dfrac{8}{5} = 16\,\mathrm{mol}$

Given: $8\,\mathrm{H_2S}$ corresponds to $5\,\mathrm{mol}$

Scaling factor = $\dfrac{5}{8}$

Final: $x = 12.8 \times \dfrac{5}{8} = 8$ $y = 16 \times \dfrac{5}{8} = 10$

Thus, $x + y = 18$.