In the electrochemical cell : Zn ,|,ZnSO_4,(0 | vedclass

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Question

(B) The cell reaction for the Daniell cell is $Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s).$Using the Nernst equation $E = E^{\circ} - \fr

Vedclass · Accepted Answer

$E_1 > E_2$

Vedclass · Answer

(B) The cell reaction for the Daniell cell is $Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s).$Using the Nernst equation $E = E^{\circ} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ where $n = 2.$For $E_1$: $[Zn^{2+}] = 0.01 \, M$ and $[Cu^{2+}] = 1.0 \, M.$$E_1 = E^{\circ} - \frac{0.059}{2} \log \frac{0.01}{1} = E^{\circ} - \frac{0.059}{2} \log(10^{-2}) = E^{\circ} + 0.059.$For $E_2$: $[Zn^{2+}] = 1.0 \, M$ and $[Cu^{2+}] = 0.01 \, M.$$E_2 = E^{\circ} - \frac{0.059}{2} \log \frac{1}{0.01} = E^{\circ} - \frac{0.059}{2} \log(10^{2}) = E^{\circ} - 0.059.$Comparing the two,$E_1 = E^{\circ} + 0.059$ and $E_2 = E^{\circ} - 0.059.$Therefore,$E_1 > E_2.$

Similar Questions

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