A strip of lead is dipped into copper sulphate solution. Use E∘(Pb2+/Pb)=−0.13 VE^\circ(\text{Pb}^{2+}/\text{Pb}) = -0.13\ \text{V}E∘(Pb2+/Pb)=−0.13 V and E∘(Cu2+/Cu)=+0.34 VE^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\ \text{V}E∘(Cu2+/Cu)=+0.34 V. What happens?
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