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.