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(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. |
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Answer» SOLUTION : (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` |
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