Use the following half-reactions to write a spontaneous reaction, calculate `E^0(cell)` for the reaction. 1)`2HClO(aq) + 2H^+(aq) + 2e^-rarrCl2(g)+2H2O(l) E^0=1.63V`  2) `Pt^2+(aq) +...

Use the following half-reactions to write a spontaneous reaction, calculate `E^0(cell)` for the reaction.

1)`2HClO(aq) + 2H^+(aq) + 2e^-rarrCl2(g)+2H2O(l) E^0=1.63V`

 2) `Pt^2+(aq) + 2e^-rarrPt(s)(E^0=1.20V)`

 3) `PbSO4(s) +2e^- rarr Pb(s) +SO4^2-(aq) (E^0=-0.31V)`

 

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These are standard reduction potential values. Higher the value, greater is the propensity of the species to get reduced. Lower the value, greater is the propensity of the species to get oxidized.

Among the given species``, the system at 1) has the highest `E^0(red)` value. Similarly, among the systems at 2) and 3), the latter has greater propensity to get oxidized, as it has the lowest `E^0(red)` value.

Therefore, the most favourable spontaneous reaction will be between 1) and 3) in the direction as described, i.e. between the following half-reactions:

`1) 2HClO(aq.)+2H^+(aq.)+ 2e^-rarrCl2(g)+2H2O(l)(E^0=1.63V)`

`3) PbSO4(s)+2e^-rarr Pb(s)+SO4^2-(aq.) (E^0=-0.31 V)`

`rArr Pb(s) +SO4^2-(aq.)rarr PbSO4(s)+2e^- (E^0=0.31 V)`

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Overall, the spontaneous reaction is:`2HClO(aq.) + 2H^+(aq.) + 2e^- + Pb(s) +SO4^2-(aq.) rarr Cl2(g)+2H2O(l)+ PbSO4(s) +2e^-`

Upon simplifying, 

`Pb(s) +SO4^2-(aq.) + 2HClO(aq.) + 2H^+(aq.) rarr PbSO4(s)+ Cl2(g)+ 2H2O(l)`

`E^0(overall) = 1.63+0.31=1.94 V`

Sources:

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