JEE Main · 2024 · Shift-IImediumEC-041

For the electrochemical cell, if E°(M2+/M) = 0.46\ V and E°(X/X2-) = 0.34\ V. Which of the following is correct?

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

For the electrochemical cell, if E°(M2+/M)=0.46 VE°_{(\mathrm{M^{2+}/M})} = 0.46\ \mathrm{V} and E°(X/X2−)=0.34 VE°_{(\mathrm{X/X^{2-}})} = 0.34\ \mathrm{V}.

Which of the following is correct?

Options
  1. a

    M+X→M2++X2−\mathrm{M + X \rightarrow M^{2+} + X^{2-}} is a spontaneous reaction

  2. b

    Ecell=0.80 VE_{\text{cell}} = 0.80\ \mathrm{V}

  3. c

    Ecell=−0.80 VE_{\text{cell}} = -0.80\ \mathrm{V}

  4. d

    M2++X2−→M+X\mathrm{M^{2+} + X^{2-} \rightarrow M + X} is a spontaneous reaction

    ✓
Correct Answerd

M2++X2−→M+X\mathrm{M^{2+} + X^{2-} \rightarrow M + X} is a spontaneous reaction

Detailed Solution

Strategy: For an electrochemical cell to be spontaneous, the standard cell potential (Ecell∘E^\circ_{\text{cell}}) must be positive. This is achieved when the half-cell with the higher reduction potential acts as the cathode.

Step 1: Compare reduction potentials

  • E\ceM2+/M∘=+0.46 VE^\circ_{\ce{M^{2+}/M}} = +0.46\text{ V}
  • E\ceX/X2−∘=+0.34 VE^\circ_{\ce{X/X^{2-}}} = +0.34\text{ V}

Step 2: Determine cathode and anode for spontaneity Cathode (Higher E∘E^\circ) = \ceM2+/M\ce{M^{2+}/M} Anode (Lower E∘E^\circ) = \ceX/X2−\ce{X/X^{2-}} Ecell∘=Ecathode∘−Eanode∘=0.46−0.34=+0.12 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.46 - 0.34 = +0.12\text{ V}

Step 3: Write the spontaneous reaction Reduction at Cathode: \ceM2++2e−→M\ce{M^{2+} + 2e^- \rightarrow M} Oxidation at Anode: \ceX2−→X+2e−\ce{X^{2-} \rightarrow X + 2e^-} Net Spontaneous Reaction: \ceM2++X2−→M+X\ce{M^{2+} + X^{2-} \rightarrow M + X}

Answer: (d) \ceM2++X2−→M+X is a spontaneous reaction\boxed{\text{Answer: (d) } \ce{M^{2+} + X^{2-} \rightarrow M + X} \text{ is a spontaneous reaction}}

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For the electrochemical cell, if E°(M2+/M) = 0.46\ V and E°(X/X2-) = 0.34\ V. Which of the… (JEE Main 2024) | Canvas Classes