MHTCET 18 Apr 2026 Shift 2 Paper

Concept:For a spontaneous cell reaction, $E_{\text{cell}}^\circ$ must be positive, meaning the cathode half-cell must have a higher reduction potential than the anode.Explanation:In the reaction, X is oxidised to $\mathrm{X}^{2+}$ at the anode, while $\mathrm{Y}^{2+}$ is reduced to Y at the cathode.Therefore, $E_{\text{cell}}^\circ = E_{\mathrm{Y}^{2+}/\mathrm{Y}}^\circ - E_{\mathrm{X}^{2+}/\mathrm{X}}^\circ$.For spontaneity, $E_{\text{cell}}^\circ > 0$, which requires $E_{\mathrm{Y}^{2+}/\mathrm{Y}}^\circ > E_{\mathrm{X}^{2+}/\mathrm{X}}^\circ$.Given values are $E_{\mathrm{Zn}^{2+}/\mathrm{Zn}}^\circ = -0.76\ \mathrm{V}$, $E_{\mathrm{Ni}^{2+}/\mathrm{Ni}}^\circ = -0.23\ \mathrm{V}$, and $E_{\mathrm{Fe}^{2+}/\mathrm{Fe}}^\circ = -0.44\ \mathrm{V}$.The decreasing order of reduction potential is Ni > Fe > Zn, since $-0.23 > -0.44 > -0.76$.Hence, Y must be the metal with the higher reduction potential, and X must be the one with the lower value.This gives X = Zn and Y = Ni.Verification: $E_{\text{cell}}^\circ = (-0.23) - (-0.76) = +0.53\ \mathrm{V}$, which is positive.Thus, the reaction is spontaneous.Answer:Option D: X = Zn, Y = Ni