derive the nernst eqn for daniel cell

Asked by aswathy sreekumar | 13th May, 2014, 08:19: PM

Expert Answer:

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, Ecell = 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

Ecell = E0cell - RT/2F ln [Zn2+]/[Cu2+]

Ecell = E?cell2.303RT/2F ln [Zn2+]/[Cu2+]

Answered by Prachi Sawant | 14th May, 2014, 10:44: AM