Question
The following galvanic cell: Zn|Zn(NO3)2 (100 ml, 1 M)||Cu(NO3)2 (100 ml, 1 M)|Cu was operated as an electrolytic cell as Cu as the anode and Zn as the cathode. A current of 0.4825 A was passed for 10 h and then the cell was allowed to function as galvanic cell. What would be the final EMF of the cell at 25°C? Assume that the only electrode reactions occurring were those involving Cu|Cu2+ and Zn|Zn2+.
Given: $E_{Cu^{2 +}|Cu}^{o}$= +0.34 V, $E_{Zn^{2 +}|Zn}^{o}$= −0.76 V, 2.303 RT/F = 0.06, log 1.9 = 0.28)
Options
Solution
Equivalent of charge used = $\frac{0.4825 \times 10 \times 3600}{96500} = 0.18\text{ F}$
Cell reaction during charge:
$Cu + Zn^{2 +} \rightarrow Cu^{2 +} + Zn $$${\frac{100 \times 1}{1000} = 0.1\text{ mole}\frac{100 \times 1}{1000} = 0.1\text{ mole} }{\text{= 0.2 eq= 0.2 eq} }{\text{Final}0.2 - 0.18 = 0.2 + 0.18 }{= 0.2\text{eq= 0.38 eq} }{\text{= 0.01 mole= 0.19 mole} }{\therefore\text{Final}\left\lbrack Zn^{2 +} \right\rbrack = \frac{0.01}{100} \times 1000 = 0.1\text{ M and }\left\lbrack Cu^{2 +} \right\rbrack = \frac{0.19}{100} \times 1000 = 1.9\text{ M}}$$
Now, cell reaction as galvanic cell:
$Zn + Cu^{2 +} \rightarrow Zn^{2 +} + Cu $$$\text{and }E_{cell} = E_{cell}^{o} - \frac{0.06}{n}.\log\frac{\left\lbrack Zn^{2 +} \right\rbrack}{\left\lbrack Cu^{2 +} \right\rbrack} = \left\lbrack 0.34 - ( - 0.76) \right\rbrack - \frac{0.06}{2}.\log\frac{0.1}{1.9} = 1.1084\text{ V}$$
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