(i) What type of a battery is lead storage battery ? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery. (ii) Calculate the potential for half-cell containing:
0.10 M `K_(2)Cr_(2)O_(7) (aq) , 0.20 M Cr^(3+) (aq) ` and `1.0 xx 10^(-4) M H^(+)` (aq)
The half-cell reaction is:
`Cr_(2)O_(7)^(2-) (aq) + 14H^(+)(aq) + 6e^(-) to 2Cr^(3+)(aq) + 7H_(2)O (l)`
and the standard electrode potential is given as `E^(@) =1.33 V`.

Explanation

(i) Lead storage battery is a secondary cell. A secondary cell after use can be recharged by passing current through it in the opposite direction. It consists of a lead anode and a grid of lead packed with lead dioxide (`PbO_2`) as cathode. A 38% solution of sulphuric acid is used as an electrolyte.
Reaction:
Anode: `Pb(s) + SO_(4)^(2-) (aq) to PbSO_(4) (s) + 2e^(-)`
Cathode: `PbO_(2)(s) + SO_(4)^(2-) (aq) + 4H^(+) (aq) + 2e^(-) to PbSO_(4) (s) + 2H_(2)O(l)`
Overall reaction: `PbO_(2)(s) +SO_(4)^(2-) (aq) + 4H^(+) (aq) + 2e^(-) to 2PbSO_(4) (s) + 2H_(2)O (l)`
(ii) The relation between E and E° is given as under :
`E =E^(@) -(RT)/(nF) ln ([C][D])/([A][B])`
For the given reaction it can be written as:
`E =E^(@) -(RT)/(nF) ln ([Cr^(3+)]^(2))/([Cr_(2)O_(7)^(2-)][H^(+)]`
Substituting the values, we have
`E = 1.33 -0.059/6 log ([0.2])^(2)/([0.1][1 xx 10^(-4)]^(14)) = 1.33 -0.059/6 log (0.2 xx 0.2)/(0.1 xx 10^(-56))`
`=1.33 -0.55 = 0.78 V`