derive the nernst eqn for daniel cell
Asked by aswathy sreekumar | 13th May, 2014, 08:19: PM
For Daniel cell,
Zn(s)+Cu2+(aq) <=> Zn2+(aq) + Cu(s)
In Daniell cell, the electrode potential for any given concentration of Cu2+ and Zn2+ ions, we write
E(Cu2+/Cu) = E0(Cu2+/Cu) – RT/2F ln(1/[Cu2+(aq)]
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?cell − 2.303RT/2F ln [Zn2+]/[Cu2+]
Answered by Prachi Sawant | 14th May, 2014, 10:44: AM
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