The electrochemical cell: $Z n / Z n S O_{4} (0.01 M) / / C u S O_{4} (1 M) / C u$
The emf of the cell is E1. When the concentration of $Z n S O_{4}$ Is changed to 1.0M and the of $C u S O_{4}$ Changed to 0.01M, the emf changed to $E_{2}$ . The relationship between $E_{1}$ and $E_{2}$ Is (Given RT/F=0.059)

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$E_{1} < E_{2}$

$E_{1} > E_{2}$

$2 E_{1} = E_{2}$

$E_{1} = E_{2}$

answer is C.

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Detailed Solution

When the conc. Of electrolyte in the anode compartment increases and in the cathode compartment decreases, the emf of the cell decreases.
$E_{c e l l}^{} = E_{c e l l}^{o} - \frac{0.059}{2} \log \frac{[Z n^{+ 2}]}{[C u^{+ 2}]}$

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