The e.m.f of the cell is: Mg|Mg2+(1 M)||Pb2+(1 M)|Pb [E0(Pb/Pb2+)=0. – InterviewSolution

1.	The e.m.f of the cell is: Mg\|Mg2+(1 M)\|\|Pb2+(1 M)\|Pb [E0(Pb/Pb2+)=0.14 V, E0(Mg2+/Mg)=−2.37 V]
Answer» The e.m.f of the cell is: $M g \| M g^{2 +} (1 M) \| \| P b^{2 +} (1 M) \| P b$ $[E_{(P b / P b^{2 +})}^{0} = 0.14 V, E_{(M g^{2 +} / M g)}^{0} = - 2.37 V]$

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