GivenCuI(s)+e−⟶Cu(s)+I−(aq);E∘=−0.188V ...(i)Cu2+(aq)+I−+e−⟶CuI(s);E∘=0.868V ...(ii)Ksp(CuI)=10−12M2 ...(iii)298R ln 10F=0.059List -1 contains half cells and List-2 contains emf values of half cellsList - IList - II(P)Cu+(aq)+e−⟶Cu(s);E01=(1)0.34 V(Q)Cu2+(aq)+e−⟶Cu+(aq);E02=(2)0.16 V(R)Cu2+(aq)+2e−⟶Cu(s);E03=(3)0.52 V(S)Cu2+(1M)+I−(10−12M)+e−⟶CuI(s);E4=(4)(E01−E03−0.02)V(5)2×E02+0.02Match the emf of cell in LIST-I with one or more values in LIST-II and choose the correct option.

GivenCuI(s)+e−⟶Cu(s)+I−(aq);E∘=−0.188V ...(i)Cu2+(aq)+I−+e

1.	GivenCuI(s)+e−⟶Cu(s)+I−(aq);E∘=−0.188V ...(i)Cu2+(aq)+I−+e−⟶CuI(s);E∘=0.868V ...(ii)Ksp(CuI)=10−12M2 ...(iii)298R ln 10F=0.059List -1 contains half cells and List-2 contains emf values of half cellsList - IList - II(P)Cu+(aq)+e−⟶Cu(s);E01=(1)0.34 V(Q)Cu2+(aq)+e−⟶Cu+(aq);E02=(2)0.16 V(R)Cu2+(aq)+2e−⟶Cu(s);E03=(3)0.52 V(S)Cu2+(1M)+I−(10−12M)+e−⟶CuI(s);E4=(4)(E01−E03−0.02)V(5)2×E02+0.02Match the emf of cell in LIST-I with one or more values in LIST-II and choose the correct option.
Answer» Given $C u I_{(s)} + e^{-} ⟶ C u_{(s)} + I_{(a q)}^{-}; E^{\circ} = - 0.188 V . . . (i) C u_{(a q)}^{2 +} + I^{-} + e^{-} ⟶ C u I_{(s)}; E^{\circ} = 0.868 V . . . (i i) K_{s p} (C u I) = 10^{- 12} M^{2} . . . (i i i)$ $\frac{298 R l n 10}{F} = 0.059$ List -1 contains half cells and List-2 contains emf values of half cells $\begin{matrix} List - I & List - II (P) & C u_{(a q)}^{+} + e^{-} ⟶ C u_{(s)}; E_{1}^{0} = & (1) & 0.34 V (Q) & C u_{(a q)}^{2 +} + e^{-} ⟶ C u_{(a q)}^{+}; E_{2}^{0} = & (2) & 0.16 V (R) & C u_{(a q)}^{2 +} + 2 e^{-} ⟶ C u_{(s)}; E_{3}^{0} = & (3) & 0.52 V (S) & C u_{(1 M)}^{2 +} + I_{(10^{- 12} M)}^{-} + e^{-} ⟶ C u I (s); E_{4} = & (4) & (E_{1}^{0} - E_{3}^{0} - 0.02) V (5) & 2 \times E_{2}^{0} + 0.02 \end{matrix}$ Match the emf of cell in LIST-I with one or more values in LIST-II and choose the correct option.