For the cell $Z n | Z n^{2 +} | | C u^{2 +} | C u,$ if the conc. Of $Z n^{2 +}$ and $C u^{2 +}$ ions is doubled ,the EMF of the cell:

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Reduces to half

Doubles

Remains same

Becomes zero

answer is C.

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Detailed Solution

$\begin{array}{l} E = E^{0} - \frac{0.059}{n F} l o g \frac{[C u^{2 +}]}{[Z n^{2 +}]} \\ = - \frac{0.059}{n F} \log \frac{2 [C u^{2 +}]}{2 [Z n^{2 +}]} \\ {=E}^{0} - \frac{0.059}{n F} l o g \frac{[C u^{2 +}]}{[Z n^{2 +}]} \end{array}$

i.e remains same

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