Question

Calculate the e.m.f. of a copper-zinc voltaic cell working with the following electrolytes 0.001 M CuSO4 and M ZnSO4.?

Answer 1

Answer on Question #49800 - Chemistry - Other

Question

Calculate the e.m.f. of a copper-zinc voltaic cell working with the following electrolytes 0.001 M $\mathrm{CuSO}_4$ and M $\mathrm{ZnSO}_4$.

Answer:

The half-cell reactions and standard potentials are:

$$\mathrm{Zn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s}) \quad E^{*} = -0.76\ \mathrm{V}\ (\text{anode})$$ $$\mathrm{Cu}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \quad E^{*} = 0.34\ \mathrm{V}\ (\text{cathode})$$

Calculate the potentials for given concentrations:

$$E = E_{Me}^{0} + (0,059 / z) \cdot \lg [Me^{z+}]$$ $$E(\mathrm{Zn}) = E_{\mathrm{Zn}}^{0} + (0,059 / z) \cdot \lg [\mathrm{Zn}^{z+}] = -0.76 + (0,059 / 2) \cdot \lg [0.001] = -0.849\ \mathrm{V}$$ $$E(\mathrm{Cu}) = E_{\mathrm{Cu}}^{0} + (0,059 / z) \cdot \lg [\mathrm{Cu}^{z+}] = 0.34 + (0,059 / 2) \cdot \lg [0.001] = 0.252\ \mathrm{V}$$

The e.m.f is:

$$E^{*}(\text{cell}) = E(\mathrm{Cu}) - E(\mathrm{Zn}) = 0.252 - (-0.849) = 1.101\ \mathrm{V}$$

Answer: 1.1 V

https://www.AssignmentExpert.com