Write the half cell reaction and the overall cells reaction for the electrochemical cell :
`Zn|Zn^(2+)(1.0M)||Pb^(2+)(1.0M)|Pb`
Calculate the standard e.m.f for the cell if standard electrode potentials (reduction ) for `Pb^(2+)|Pb and Zn^(2+)|Zn` electrodes are -0.126 V and -0.763 V respectively.

Explanation

Zn electrode acts as anode while Pb electrode acts as cathode and , therefore , oxidation occurs at zinc electrode and reduction occurs at lead electrode . The half cell reactions are :
Oxidation half reactions :
`Zn(s) rarr Zn^(2+)(aq)+2e^(-)`
Reduction half reaction :
`Pb^(2+)(aq) +2e^(-) rarr Pb(s)`
Overall cell reaction
`Zn(s) +Pb^(2+)(aq) rarrZn^(2+) (aq) +Pb(s)`
E.M.F = `E_R^@ - E_L^@`
`=-0.126-(-0.763)`
= 0.637 V.
Note . The EMF of the cell can also simply be calculated as :
`E_("cell")^@=E_R^@-E_L^@`
If the value of `E_("cell")^@ ` is positive , then the cell is correctly represented , i.e. , oxidation occurs at left electrode (anode) and reduction occurs at the right electrode (cathode). On the other hand , if the value of `E_("cell")^@` comes out to be negative , then the cells is wrongly represented . Its order is reversed to get positive value for `E_("cell")^@`.