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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2026 · 5 April · Evening Shift · Chemistry · Electrochemistry

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

One half cell in a voltaic cell is constructed by dipping silver rod in AgNO3AgNO_3 solution of unknown concentration, other half cell is ZnZn rod dipped in 11 molar solution of ZnSO4ZnSO_4. A voltage of 1.60 V1.60\text{ V} is measured at 298 K298\text{ K} for this cell. What is the concentration of Ag+Ag^+ ions used in terms of log⁡x\log x (x=[Ag+]x = [Ag^+]) ?
EZn2+/Zn∘=−0.76 VE^\circ_{Zn^{2+}/Zn} = -0.76\text{ V}, EAg+/Ag∘=+0.80 VE^\circ_{Ag^+/Ag} = +0.80\text{ V}, 2.303RTF=0.059 V\dfrac{2.303 RT}{F} = 0.059\text{ V}

Options

  1. A23.9\dfrac{2}{3.9}
  2. B45.9\dfrac{4}{5.9}
  3. C2.92\dfrac{2.9}{2}
  4. D5.94\dfrac{5.9}{4}

Correct answer

B. 45.9\dfrac{4}{5.9}

Step-by-step solution

The cell reaction is Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s)Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)

The standard cell potential is given by:
Ecell∘=Ecathode∘−Eanode∘=0.80−(−0.76)=1.56 VE^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.76) = 1.56\text{ V}

Using the Nernst equation:
Ecell=Ecell∘−0.059nlog⁡[Zn2+][Ag+]2E_{cell} = E^\circ_{cell} - \dfrac{0.059}{n} \log \dfrac{[Zn^{2+}]}{[Ag^+]^2}

Substituting the given values (n=2n = 2, [Zn2+]=1 M[Zn^{2+}] = 1\text{ M}, Ecell=1.60 VE_{cell} = 1.60\text{ V}):
1.60=1.56−0.0592log⁡1x21.60 = 1.56 - \dfrac{0.059}{2} \log \dfrac{1}{x^2}

0.04=−0.0592(−2log⁡x)0.04 = - \dfrac{0.059}{2} (-2 \log x)

0.04=0.059log⁡x0.04 = 0.059 \log x

log⁡x=0.040.059=45.9\log x = \dfrac{0.04}{0.059} = \dfrac{4}{5.9}

Answer: 45.9\dfrac{4}{5.9}
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