The emf (in V) of a Daniell cell containing 0 | vedclass

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(D) The standard cell potential is calculated as: $E_{	ext{cell}}^{\circ} = E_{Cu^{2+} / Cu}^{\circ} - E_{Zn^{2+} / Zn}^{\circ} = 0.34 - (-0.76) = 1.

Vedclass · Accepted Answer

$1.07$

Vedclass · Answer

(D) The standard cell potential is calculated as: $E_{	ext{cell}}^{\circ} = E_{Cu^{2+} / Cu}^{\circ} - E_{Zn^{2+} / Zn}^{\circ} = 0.34 - (-0.76) = 1.1 \ V$. Using the $Nernst$ equation: $E_{	ext{cell}} = E_{	ext{cell}}^{\circ} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$. For the reaction $Zn + Cu^{2+} ightarrow Zn^{2+} + Cu$,the number of electrons transferred is $n = 2$. Substituting the values: $E_{	ext{cell}} = 1.1 - \frac{0.059}{2} \log \frac{0.1}{0.01} = 1.1 - 0.0295 	imes \log(10)$. Since $\log(10) = 1$,we get $E_{	ext{cell}} = 1.1 - 0.0295 = 1.0705 \ V \approx 1.07 \ V$.

The $emf$ (in $V$) of a $Daniell$ cell containing $0.1 \ M \ ZnSO_4$ and $0.01 \ M \ CuSO_4$ solutions at their respective electrodes is $\left(E_{Cu^{2+} / Cu}^{\circ}=+0.34 \ V ; E_{Zn^{2+} / Zn}^{\circ}=-0.76 \ V\right)$

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