Question

The half reactions that occur in a lead acid battery are:
`PbSO_(3)(s)+2e^(-)toPb(s)+SO_(4)^(2-)(aq)E^(@)=-0.36V`
`PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l)E^(@)=+1.69V`
Calculate the overall potential for the cell in discharging reaction, `E_(cell)^(@)` Give answer in nearest single digit integer.

Answer 1

Correct Answer - 2
Use the `E^(@)` values to decide the direction of the overall cell reaction.
the `E^(@)` value for the reduction of `PbO_(2)` is more positive than that for the reduction of `PbSO_(4)`, so the `PbO_(2)` half reaction will oxidize `Pb` to `PbSO_(4)`.
Anode `Pb(S)+SO_(4)^(2-(aq)toPbSO_(4)(s)+2e^(-)`
Cathode
`PbO_(2)(s)+4H(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(I)`
`E_(cell)^(@)=(E_("cathode")^(@))_(RP)-(E_("anode")^(@))_(RP)`
`=(+1.69V)-(-0.36V)=+2.05V`