JEE Main 2025 Jan 23 Shift 1, Chemistry Q5: Electrochemical Cell And Nernst Equation

JEE Main2025Jan 23, Shift 1Chemistry

$F e O_{4}^{2 -} + 2.0 V F e^{3 +} 0.8 V F e^{2 +} - 0.5 V F e^{0}$

In the above diagram, the standard electrode potentials are given in volts (over the arrow).

The value of $E_{F e O_{4}^{2 -} / F e^{2 +}}^{⊖}$ is

Solution

We combine two half-reactions to get $F e O_{4}^{2 -} \to F e^{2 +}$ :

Step 1: $F e O_{4}^{2 -} \to F e^{3 +}$ involves 3 electrons, $E_{1}^{\circ} = 2.0$ V.

Step 2: $F e^{3 +} \to F e^{2 +}$ involves 1 electron, $E_{2}^{\circ} = 0.8$ V.

Net process: $F e O_{4}^{2 -} \to F e^{2 +}$ involves $n = 4$ electrons.

Use $n_{n e t} E_{n e t}^{\circ} = n_{1} E_{1}^{\circ} + n_{2} E_{2}^{\circ}$ (additive in $Δ G^{\circ}$ ):

$4 \cdot E^{\circ} = 3 (2.0) + 1 (0.8) = 6.8$

$∴ E_{F e O_{4}^{2 -} / F e^{2 +}}^{\circ} = \frac{6.8}{4} = 1.7$ V.

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