Chemistry · Electrochemistry

JEE Main 2026 — 23 January, Evening Shift — Question 48

Consider the above electrochemical cell where a metal electrode ( M ) is undergoing redox reaction by forming M+(M→M++e)\mathrm{M}^{+}\left(\mathrm{M} \rightarrow \mathrm{M}^{+}+\mathrm{e}\right). The cation M+\mathrm{M}^{+}is present in two different concentrations c1\mathrm{c}_{1} and c2\mathrm{c}_{2} as shown above. Which of the following statement is correct for generating a positive cell potential?

Question figure
  1. Option A:

    If c1\mathrm{c}_{1} is present at anode, then c1=c2\mathrm{c}_{1}=\mathrm{c}_{2}

  2. Option B:

    If c1\mathrm{c}_{1} is present at cathode, then c1<c2\mathrm{c}_{1}<\mathrm{c}_{2}

    Correct
  3. Option C:

    If c1\mathrm{c}_{1} is present at cathode, then c1>c2\mathrm{c}_{1}>\mathrm{c}_{2}

  4. Option D:

    If c1\mathrm{c}_{1} is present at anode, then c1>c2\mathrm{c}_{1}>\mathrm{c}_{2}

Answer: B

Step-by-step solution

(1) If C1\mathrm{C}_{1} is at anode ⇒ cell reaction M+(C2)→M+(C1)\mathrm{M}^{+}\left(\mathrm{C}_{2}\right) \rightarrow \mathrm{M}^{+}\left(\mathrm{C}_{1}\right) Ecell =−0.059log⁡C1C2\mathrm{E}_{\text {cell }}=-0.059 \log \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}} ∴Ecell>0⇒C1<C2\therefore \mathrm{E}_{\mathrm{cell}}>0 \Rightarrow \mathrm{C}_{1}<\mathrm{C}_{2} (2) If C1\mathrm{C}_{1} is at cathode M+(C1)→M+(C2)\mathrm{M}^{+}\left(\mathrm{C}_{1}\right) \rightarrow \mathrm{M}^{+}\left(\mathrm{C}_{2}\right) Ecell =−0.059log⁡C2C1>0\mathrm{E}_{\text {cell }}=-0.059 \log \frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}>0 C2<C1\mathrm{C}_{2}<\mathrm{C}_{1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Electrochemistry
Topic
Nernst Equation and Electrochemical Series
Consider the above electrochemical cell where a metal electrode ( M )… | JEE Main 2026 PYQ with Solution · DhiX AI