In the electrochemical cell: the emf of this Daniell cell is . When the concentration of is changed to 1.0 M and that of changed to 0.01 M, the emf changes to . From the followings, which one is the relationship between and ? (Given, )

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Solution

Q.
In the electrochemical cell:

$Z n | Z n S O_{4} (0.01 M) | | C u S O_{4} (1.0 M) | C u,$

the emf of this Daniell 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 followings, which one is the relationship between $E_{1}$ and $E_{2}$ (Given, $\frac{R T}{F} = 0.059$ ) [2017, 2003]

1 $E_{1} < E_{2}$

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

3 $E_{2} = 0 \neq E_{1}$

4 $E_{1} = E_{2}$

Ans.
(2)

$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$

$E_{1} = E ° - \frac{0.059}{2} \log \frac{0.01}{1}$

$E_{1} = E ° - \frac{0.059}{2} (- 2) = E ° + 0.059$

$E_{2} = E ° - \frac{0.059}{2} \log \frac{1}{0.01} = E ° - 0.059$

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

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