In the electrochemical cell Zn ZnSO_4(0.01M) CuSO_4(1.0M) Cu , the emf is E_1 . When the concentrations are swapped ( ZnSO_4 = 1.0M , CuSO_4 = 0.01M ) - Chemistry MCQ NEET (UG) 2017 2026 | ExamTest.live

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In the electrochemical cell Zn ∣ ZnSO_{4} (0.01 M) ∥ CuSO_{4} (1.0 M) ∣ Cu, the emf is E_{1} . When the concentrations are swapped (ZnSO_{4} = 1.0 M, CuSO_{4} = 0.01 M), the emf is E_{2} . The relationship is:

Solution & Step-by-step Explanation

Using Nernst Equation: E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.059}{2} lo g \frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]} For E_{1} : E_{1} = E^{\circ} - \frac{0.059}{2} lo g \frac{0.01}{1} = E^{\circ} - \frac{0.059}{2} (- 2) = E^{\circ} + 0.059 .For E_{2} : E_{2} = E^{\circ} - \frac{0.059}{2} lo g \frac{1}{0.01} = E^{\circ} - \frac{0.059}{2} (2) = E^{\circ} - 0.059 .Clearly, E_{1} > E_{2} .

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In the electrochemical cell Zn ∣ ZnSO_{4} (0.01 M) ∥ CuSO_{4} (1.0 M) ∣ Cu, the emf is E_{1} . When the concentrations are swapped (ZnSO_{4} = 1.0 M, CuSO_{4} = 0.01 M), the emf is E_{2} . The relationship is:

E_{1} < E_{2}

E_{1} > E_{2}

E_{2} = 0 \neq = E_{1}

E_{1} = E_{2}

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