JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Redox Reactions and Electrochemistry

One half cell in a voltaic cell is constructed by dipping silver rod in

AgNO_{3}

solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of

ZnSO_{4}

.
A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of

Ag^{+}

ions used in terms of

lo g x

(

x = [Ag^{+}]

E_{Zn^{2 +} / Zn}^{Θ} = - 0.76 V, E_{Ag^{+} / Ag}^{Θ} = + 0.80 V

\frac{2.303 RT}{F} = 0.059 V

Answer: (B)

Solution

Zinc is the anode and silver the cathode, so

E_{cell}^{⊖} = 0.80 - (- 0.76) = 1.56 V

for

Zn + 2A g^{+} \to Z n^{2 +} + 2Ag

with

n = 2

. The Nernst equation gives

E_{cell} = E_{cell}^{⊖} - \frac{0.059}{2} lo g \frac{[ Z n ^{2 +} ]}{[ A g ^{+} ] ^{2}}

. With

[Z n^{2 +}] = 1

and

E_{cell} = 1.60 V

1.60 = 1.56 - \frac{0.059}{2} lo g \frac{1}{x ^{2}} = 1.56 + 0.059 lo g x

, so

0.04 = 0.059 lo g x

. The concentration of silver ions satisfies

lo g x = \frac{4}{5.9}

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option