Answer:
- 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.
