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.

Write the half cell reaction and the overall cells reaction for the el

1.	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.
Answer» Solution :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")^@`.