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The reduction potential of the two half ...

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

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
`Pb+SO_(4)^(2-)+2Ag^(+)(aq)rarr2Ag(s)+PbSO_(4)`
`Emf=E_(c)-E_(a)=0.80-(-0.31)`
`=0.80+0.31=1.11V`
As emf is positive so the cell reac tion is spontaneous .
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