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Question Paper from: IIT JEE Main 2021 TestSeries Paper 2

In the electrochemical cell:

Zn $∣$ ZnSO₄ (0.01 M) $∥$ CuSO₄ (1.0 M) $∣$ Cu, the emf of this daniel cell is E1. When the concentration of ZnSO₄ is changed to 1.0M and that of CuSO₄ changed to 0.01M, the emf changes to E_2. Which one is the relationship between E1 and E2 from the following? (Given $\frac{RT}{F} = 0.059$ )

E₁ = E₂

E₂ = 0 $\neq =$ E₁

E₁ < E₂

E₁ > E₂

Solution:

Chapter Name: Electro Chemistry

Difficulty Level: Moderate

Cell Notation : Zn $∣$ ZnSO₄ (0.01 M) $∥$ CuSO₄ (1.0 M) $∣$ Cu

$∴ E_{1} = E_{ce ll}^{o} - \frac{2.303 RT}{2 F} \times l o g \frac{0.01}{1}$

$E_{2} = E_{ce ll}^{o} - \frac{2.303 RT}{2 F} \times l o g \frac{1}{0.01}$

$\frac{0.01}{1} \times \frac{0.01}{1} = \frac{0.01}{100}$

$∴ E_{1} > E_{2}$

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IIT JEE Main 2021 TestSeries Paper 2

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