Sn(s) | Sn2+ || Pb2+ | Pb(s) Sn2+ +2e- <=>Sn e°=-0,14V Pb2+ + 2e- <=> Pb e° = -0,13V what is E°?

Method 1 :

e°[Sn²⁺(aq)|Sn(s)] = -0,14 V
e°[Pb²⁺(aq)|Pb(s)] = -0,13 V

E°[Sn(s)|Sn²⁺(aq)||Pb²⁺(aq)|Pb(s)]
= e°[Pb²⁺(aq)|Pb(s)] - e°[Sn²⁺(aq)|Sn(s)]
= (-0,13 V) - (-0,14 V)
= +0,01 V


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Method 2 :

The cell overall reaction represented by the cell diagram Sn(s)|Sn²⁺(aq)||Pb²⁺(aq)|Pb(s) is :
Sn(s) + Pb²⁺(aq) → Sn²⁺(aq) + Pb(s) ...... E°

Sn(s) → Sn²⁺(aq) + 2e⁻ ...... e° = +0,14 V
Pb²⁺(aq) + 2e⁻ → Pb(s) ...... e° = -0,13 V

Add the above two electrochemical equations, and cancel 2e⁻ on the both sides.
Sn(s) + Pb²⁺(aq) → Sn²⁺(aq) + Pb(s) ...... E° = (+0.14 V) - (0,13V) = +0,01 V