What is the value (in v) of eºcell for the following reaction? zn2+ (aq) + pb (s) → zn (s) + pb2+ (aq)

- 0.63 V.

Explanation:

From the question , the reaction is as follows -

Zn²⁺ (aq) + Pb (s)   →   Zn (s) + Pb²⁺ (aq)

From the above reaction ,

Zn²⁺ is reduced to Zn , as the oxidation state changes from +2 to 0

and ,

Pb is oxidized to Pb²⁺ , as the oxidation state changes from  0 to +2 .

In a cell , the process of oxidation takes place in the Anode , and reduction takes place in Cathode .

Hence , the half cell reaction taking place at anode and cathode is as follows -

Cathode reaction:  Zn²⁺ (aq) +  2e⁻ → Zn (s) ; E⁰ Zn²⁺ /Zn = - 0.76 V.

Anode reaction: Pb (s) → Pb²⁺ (aq) +  2e⁻  ;   E⁰ Pb²⁺ /Pb = - 0.13 V

The  E⁰cell is calculated as the difference in the  E⁰cathode and  E⁰anode .

E⁰cell =  E⁰cathode -  E⁰anode = - 0.76 - (-0.13) = - 0.63 V.

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