MHTCET 15 Apr 2026 Shift 2 Paper

Concept:The standard cell potential is the difference between the cathode and anode standard reduction potentials.Explanation:In the given cell notation, $\mathrm{Al}$ is the anode and $\mathrm{Pb}$ is the cathode.So, oxidation occurs at aluminium and reduction occurs at lead ion.The cell equation is:$$E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0$$Given: $E_{\text{cell}}^0 = 1.5\ \mathrm{V}$ and $E_{\text{Pb}^{2+}/\mathrm{Pb}}^0 = -0.14\ \mathrm{V}$.Let $E_{\text{Al}^{3+}/\mathrm{Al}}^0 = E_{\mathrm{Al}}^0$.Substituting the values:$$1.5 = -0.14 - E_{\mathrm{Al}}^0$$Therefore:$$E_{\mathrm{Al}}^0 = -0.14 - 1.5 = -1.64\ \mathrm{V}$$Answer:The correct option is B, i.e., $-1.64\ \mathrm{V}$.