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Chemistry · Electrochemistry

JEE Main 2026 — 5 April, Evening Shift — Question 55

One half cell in a voltaic cell is constructed by dipping silver rod in AgNO3\mathrm{AgNO}_{3} solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO4\mathrm{ZnSO}_{4}. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag+\mathrm{Ag}^{+}ions used in terms of log⁡x(x=[Ag+])\log \mathrm{x}\left(\mathrm{x}=\left[\mathrm{Ag}^{+}\right]\right)? EZn2+/ZnΘ=−0.76 V,EAg+/AgΘ=+0.80 V\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\Theta}=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\Theta}=+0.80 \mathrm{~V}, 2.303RTF=0.059 V\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}

  1. Option A:

    23.9\frac{2}{3.9}

  2. Option B:

    45.9\frac{4}{5.9}

    Correct
  3. Option C:

    2.92\frac{2.9}{2}

  4. Option D:

    5.94\frac{5.9}{4}

Answer: B

Step-by-step solution

Cell Reaction:

Zn(s)+2Ag+(aq)⟶2Ag(s)+Zn2+(aq)Ecell=Ecell∘−0.059nlog⁡(Q)1.6=1.56−0.0592log⁡(1[Ag+]2)0.04=0.059log⁡[Ag+]log⁡[Ag+]=45.9\begin{aligned} \mathrm{Zn(s)}+2\mathrm{Ag}^{+}(\mathrm{aq}) &\longrightarrow 2\mathrm{Ag(s)}+\mathrm{Zn}^{2+}(\mathrm{aq})\\ E_{\mathrm{cell}} &=E_{\mathrm{cell}}^{\circ}-\frac{0.059}{n}\log(Q)\\ 1.6 &=1.56-\frac{0.059}{2}\log\left(\frac{1}{[\mathrm{Ag}^{+}]^2}\right)\\ 0.04 &=0.059\log[\mathrm{Ag}^{+}]\\ \log[\mathrm{Ag}^{+}] &=\frac{4}{5.9} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Nernst Equation and Electrochemical Series

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