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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. |
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Answer» 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` |
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