For the electro chemical cell $\mathrm{M}\left|\mathrm{M}^{2+} \| \mathrm{X}\right| \mathrm{X}^{2-}$ If $\mathrm{E}_{\left(\mathrm{M}^{2+} / \mathrm{M}\right)}^{0}=0.46 \mathrm{~V}$ and $\mathrm{E}_{\left(\mathrm{X} / \mathrm{X}^{2-}\right)}^{0}=0.34 \mathrm{~V}$. Which of the following is correct?

(C) $\mathrm{M}^{2+}+\mathrm{X}^{2-} \rightarrow \mathrm{M}+\mathrm{X}$ is a spontaneous reaction

(A) $\mathrm{E}_{\text {cell }}=-0.80 \mathrm{~V}$

(B) $\mathrm{M}+\mathrm{X} \rightarrow \mathrm{M}^{2}+\mathrm{X}^{2-}$ is a spontaneous reaction

(D) $\mathrm{E}_{\text {cell }}=0.80 \mathrm{~V}$

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Redox Reactions and Electrochemistry

For the electro chemical cell

M M^{2 +} ∥ X X^{2 -}

E_{(M^{2 +} / M)}^{0} = 0.46 V

and

E_{(X / X^{2 -})}^{0} = 0.34 V

.
Which of the following is correct?

Answer: (C)

E_{ce l l}^{0} = E_{c a t h o d e}^{0} - E_{an o d e}^{0}

. Writing the cell as

M ∣ M^{2 +} ∥ X ∣ X^{2 -}

with M oxidised at the anode and X reduced at the cathode:

E_{ce l l}^{0} = E_{X / X^{2 -}}^{0} - E_{M^{2 +} / M}^{0} = 0.34 - 0.46 = - 0.12 V

Since this is negative, the reaction as written (

M + X \to M^{2 +} + X^{2 -}

) is non-spontaneous; the reverse,

M^{2 +} + X^{2 -} \to M + X

, is the spontaneous reaction (with

E^{0} = + 0.12

V).

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option