CBSE Class 12-science Answered

For Daniel cell,

Zn(s)+Cu²⁺(aq) <=> Zn²⁺(aq) + Cu(s)

Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu

In Daniell cell, the electrode potential for any given concentration of Cu²⁺ and Zn²⁺ ions, we write

For Cathode:

E_(Cu2+/Cu) = E⁰_(Cu2+/Cu) – RT/2F ln(1/[Cu²⁺(aq)]

For Anode:

E_{(Zn2+/Zn) =} E⁰_(Zn2+/Zn) – RT/2F ln(1/[Zn²⁺(aq)]

The cell potential, E_cell = E_(Cu2+/Cu) - E_(Zn2+/Zn)

= E⁰_(Cu2+/Cu) – RT/2F ln(1/[Cu²⁺(aq)] - E⁰_(Zn2+/Zn) + RT/2F ln(1/[Zn²⁺(aq)]

= E⁰_(Cu2+/Cu) -E⁰_(Zn2+/Zn)- RT/2F ln(1/[Cu²⁺(aq)] + RT/2F ln(1/[Zn²⁺(aq)]

= E⁰_(Cu2+/Cu) -E⁰_(Zn2+/Zn)- RT/2F (ln(1/[Cu²⁺(aq)]- ln(1/[Zn²⁺(aq)])

Therefore, Nernst equation for Daniel cell is

E_cell = E⁰_cell - RT/2F ln [Zn²⁺]/[Cu²⁺]

E_cell = E^?_cell − 2.303RT/2F ln [Zn²⁺]/[Cu²⁺]