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The reduction potential of the two half ...

The reduction potential of the two half cell reaction (occurring in an electrochemical cell) are
`PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)(E^(@)=-0.31V)`
`Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=0.80V)`
The feasible reaction will be

A

`Pb(s)+SO_(4)^(2-)(aq)+2Ag^(+)+(aq)rarr2Ag(s)+PbSO_(4)(s)`

B

`PbSO_(4)(s)+2Ag^(+)(aq)rarrPb(s)+SO_(4)^(2-)(aq)+2Ag(s)`

C

`Pb(s)+SO_(4)^(2-)(s)+Ag(s)rarrAg^(+)(aq)+PbSO_(4)(s)`

D

`PbSO_(4)(s)+Ag(s)rarrAg^(+)(aq)+Pb(s)+SO_(4)^(2-)(aq)`

Text Solution

Verified by Experts

The correct Answer is:
A
Higher the reduction potential easier to reduce,
Therefore two h alf cell reactions are:
`Pb(s)+SO_(4)^(2-)(aq)rarrPbSO_(4)(s)+2e^(-)`
`Ag^(+)(aq)+e^(-)rarrAg(s)xx2`
`Pb (s)+SO_(4)^(2-) (aq)+2Ag^(+)(aq)rarrPbSO_(4)(S)+Ag(s)`
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