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Chemistry · PYQ NEET UG 2017
हिन्दीIn the electrochemical cell, \( \text{Zn} \mid \text{ZnSO}_4(0.01 \text{ M}) \parallel \text{CuSO}_4(1.0 \text{ M}) \mid \text{Cu} \), the emf of this Daniel Cell is \( \text{E}_1 \). When the concentration of \( \text{ZnSO}_4 \) is changed to \( 1.0 \text{ M} \) and that of \( \text{CuSO}_4 \) changed to \( 0.01 \text{ M} \), the emf changes to \( \text{E}_2 \). From the following, which one is the relationship between \( \text{E}_1 \) and \( \text{E}_2 \) (Given, \( \frac{\text{RT}}{\text{F}} = 0.059 \))?
Correct answer · Option 2
\( \text{E}_1 > \text{E}_2 \)
Explanation
Using Nernst equation, \( \text{E}_1 = \text{E}^\circ - \frac{0.059}{2}\log(\frac{0.01}{1}) \) and \( \text{E}_2 = \text{E}^\circ - \frac{0.059}{2}\log(\frac{1}{0.01}) \). Thus \( \text{E}_1 > \text{E}_2 \).
