JEE Main 2026 Apr 5 Shift 2, Chemistry Q7: Electrochemical Cell And Nernst Equation

JEE Main2026Apr 5, Shift 2Chemistry

One half cell in a voltaic cell is constructed by dipping a silver rod in $A g N O_{3}$ solution of unknown concentration; the other half cell is a Zn rod dipped in 1 molar $Z n S O_{4}$ solution. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of $A g^{+}$ ions used in terms of $lo g x$ (where $x = [A g^{+}]$ )?

$E_{Z n^{2 +} / Z n}^{\circ} = - 0.76$ V, $E_{A g^{+} / A g}^{\circ} = + 0.80$ V, $\frac{2.303 R T}{F} = 0.059$ V.

A
$\frac{2}{3.9}$
B
$\frac{4}{5.9}$
C
$\frac{2.9}{2}$
D
$\frac{5.9}{4}$

Solution

Cell reaction: $Z n (s) + 2 A g^{+} (a q) \to Z n^{2 +} (a q) + 2 A g (s)$ , with $n = 2$ .

$E_{cell}^{\circ} = 0.80 - (- 0.76) = 1.56$ V.

Nernst: $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ A g ^{+} ] ^{2}}$ .

With $[Z n^{2 +}] = 1$ M: $1.60 = 1.56 - \frac{0.059}{2} lo g \frac{1}{[ A g ^{+} ] ^{2}} = 1.56 + 0.059 lo g [A g^{+}]$ .

$0.04 = 0.059 lo g [A g^{+}] \Rightarrow lo g [A g^{+}] = \frac{0.04}{0.059} = \frac{4}{5.9}$ .

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JEE Main 2026 Apr 5 Shift 2, Chemistry Q7: Electrochemical Cell And Nernst Equation

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