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EMF

Zn | ZnSO 40....

Question

$Z n | Z n S O_{4} (0.01 M) | | C u S O_{4} (1.0 M) | C u$
The emf of the cell is $E_{1}$ . When the concentration of $Z n S O_{4}$ is changed to $1.0 M$ and that of $C u S O_{4}$ changed to $0.01 M$ , the emf changes to $E_{2}$ . From the following, which one is the relationship between $E_{1}$ and $E_{2}$ ?
(Given, $\frac{R T}{F} = 0.059)$

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 C $E_{1} > E_{2}$
$Z n | Z n S O_{4} (0.01 M) | | C u S O_{4} (1.0 M) | C u$
$∴ E_{1} = E_{c e l l}^{\circ} - \frac{2.303 R T}{2 F} \times l o g \frac{(0.01)}{1}$
When concentrations are changed
$∴ E_{2} = E_{c e l l}^{\circ} - \frac{2.303 R T}{2 F} \times l o g \frac{1}{0.01}$
i.e., $E_{1} > E_{2}$

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Similar questions

Q. In the electrochemical cell:

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