JEE Main 2025 Apr 8 Shift 2, Chemistry Q25: Electrochemical Cell And Nernst Equation

JEE Main2025Apr 8, Shift 2Chemistry

Consider the following half cell reaction

$C r_{2} O_{7}^{2 -} (a q) + 6 e^{-} + 14 H^{+} (a q) \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$

The reaction was conducted with the ratio of $\frac{[ C r ^{3 +} ] ^{2}}{[ C r _{2} O _{7}^{2 -} ]} = 1 0^{- 6}$ . The pH value at which the EMF of the half cell will become zero is ________. (nearest integer value)

[Given: standard half cell reduction potential $E_{C r_{2} O_{7}^{2 -}, H^{+} / C r^{3 +}}^{\circ} = 1.33$ V, $\frac{2.303 R T}{F} = 0.059$ V]

Your answer

Solution

The Nernst equation for this half-cell is $E = E^{\circ} - \frac{0.059}{6} lo g \frac{[ C r ^{3 +} ] ^{2}}{[ C r _{2} O _{7}^{2 -} ] [ H ^{+} ] ^{14}}$ .

Setting $E = 0$ and substituting the given ratio $\frac{[ C r ^{3 +} ] ^{2}}{[ C r _{2} O _{7}^{2 -} ]} = 1 0^{- 6}$ : $0 = 1.33 - \frac{0.059}{6} lo g \frac{1 0 ^{- 6}}{[ H ^{+} ] ^{14}}$ .

Rearranging: $\frac{1.33 \times 6}{0.059} = lo g \frac{1 0 ^{- 6}}{[ H ^{+} ] ^{14}} = - 6 - 14 lo g [H^{+}]$ .

Evaluating the left side: $\frac{1.33 \times 6}{0.059} \approx 135.25$ .

So $135.25 = - 6 - 14 lo g [H^{+}]$ , which gives $14 lo g [H^{+}] = - 141.25$ , so $lo g [H^{+}] \approx - 10.09$ , i.e. pH $\approx 10.09$ .

Rounded to the nearest integer, the pH is 10.

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q25: Electrochemical Cell And Nernst Equation

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