JEE Main 2025 Jan 23 Shift 2, Chemistry Q2: Electrochemical Cell And Nernst Equation

JEE Main2025Jan 23, Shift 2Chemistry

Standard electrode potentials for a few half cells are mentioned below:

$E_{C u^{2 +} / C u}^{o} = 0.34 V, E_{Z n^{2 +} / Z n}^{o} = - 0.76 V$

$E_{A g^{+} / A g}^{o} = 0.80 V, E_{M g^{2 +} / M g}^{o} = - 2.37 V$

Which one of the following cells gives the most negative value of $Δ G^{\circ}$ ?

Solution

$Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ , so the most negative $Δ G^{\circ}$ comes from the cell with the largest positive $E_{ce l l}^{\circ}$ .

Compute $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ}$ for each option:

(1) $E^{\circ} = 0.80 - (- 0.76) = + 1.56 V$

(2) $E^{\circ} = - 2.37 - (- 0.76) = - 1.61 V$ (negative, so $Δ G^{\circ} > 0$ )

(3) $E^{\circ} = - 2.37 - 0.80 = - 3.17 V$ (negative)

(4) $E^{\circ} = 0.80 - 0.34 = + 0.46 V$

Only options (1) and (4) have positive $E^{\circ}$ . Of these, option (1) is the larger, giving the most negative $Δ G^{\circ} = - 2 F (1.56) V$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.