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The e.m.f of ...

Question

The e.m.f of a Daniell cell at $298 K$ is $E_{1}$
$Z n | Z n S O_{4} (0.01 M) | | C u S O_{4} (1.0 M) | C u$
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 $E_{1}$ and $E_{2}$ ?

A

E_{1} > E_{2}

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B

E_{1} < E_{2}

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C

E_{1} = E_{2}

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D

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

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Solution

The correct option is A $E_{1} > E_{2}$
Cell reaction can be represented by
$Z n + C u^{2 +} \to C u + Z n^{2 +}$
Applying Nernst equation,
$E_{1} = E_{c e l l}^{\circ} - \frac{0.059}{2} l o g \frac{0.01}{1}$
When the concentrations of $Z n S O_{4}$ and $C u S O_{4}$ are altered,
$E_{2} = E_{c e l l}^{\circ} - \frac{0.059}{2} l o g \frac{1}{0.01}$
So $E_{1} > E_{2}$

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Q. The emf of Daniell cell at

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