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The e.m.f of a Daniell cell at 298K is $E_{1}$ .
$Z n | \begin{matrix} Z n S O_{4} \\ (0.01 M) \end{matrix} | | \begin{matrix} C u S O_{4} \\ (1.0 M) \end{matrix} | C u$
When the concentration of $Z n S O_{4}$ is 1.0 M and that of $C u S O_{4}$ is 0.01 M, the e.m.f. changed to $E_{2}$ . What is the relationship between $E_{1} & E_{2}$ ?

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$E_{1} > E_{2}$

$E_{1} < E_{2}$

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

$E_{1} = E_{2}$

answer is B.

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Detailed Solution

Cell reaction is, $Z n + C u^{2 +} \to Z n^{2 +} + C u$

$E_{c e l l} = E_{c e l l}^{0} - \frac{R T}{n F} ℓ n \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

Greater the factor $[\frac{(Z n^{2 +})}{(C u^{2 +})}]$ , less is the EMF

Hence $E_{1} > E_{2}$

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