JEE Main 2026 Apr 4 Shift 2, Chemistry Q23: Electrochemical Cell And Nernst Equation

JEE Main2026Apr 4, Shift 2Chemistry

An electrochemical cell. consist of the following two redox couples,

$M^{x +} (a q) / M (s)$ $[E °_{r e d} = + 0.15 V]$ and

$F e^{3 +} (a q) / F e (s)$ $[E °_{r e d} = - 0.036 V]$

The cell EMF ( $E_{ce l l}$ ) is recorded to be 0.2057 V.

If the reaction quotient of the electrochemical reaction is found to be $1 0^{- 2}$ , then the value of x is ________ (Nearest integer)

[Given: M is a p-block metal and $\frac{2.303 R T}{F} = 0.059 V$ ]

Your answer

Solution

$E °_{r e d} (M^{x +} / M) = + 0.15$ V is more positive than that of Fe $^{3 +}$ /Fe, so M $^{x +}$ /M is the cathode (reduction) and Fe is the anode (oxidation).

$E °_{ce l l} = 0.15 - (- 0.036) = + 0.186$ V.

The cell reaction balances electrons: Fe loses 3 (Fe $\to$ Fe $^{3 +}$ + 3e $^{-}$ ) and M $^{x +}$ gains x. LCM gives n = 3x electrons transferred.

Nernst equation: $E_{ce l l} = E °_{ce l l} - \frac{0.059}{n} lo g Q$ .

$0.2057 = 0.186 - \frac{0.059}{3 x} lo g (1 0^{- 2}) = 0.186 - \frac{0.059}{3 x} (- 2) = 0.186 + \frac{0.118}{3 x}$ .

$\frac{0.118}{3 x} = 0.0197 \Rightarrow 3 x = \frac{0.118}{0.0197} \approx 5.99$ .

$x \approx 2$ .

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JEE Main 2026 Apr 4 Shift 2, Chemistry Q23: Electrochemical Cell And Nernst Equation

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