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The two half - cell reactions of an elec...

The two half - cell reactions of an electrochemical cell is given as -
`Ag^(+)+e rarr Ag, E_(Ag^(+)|Ag)^(@)=-0.3995V`
`Fe^(2+)rarr Fe^(3+)+e, E_(Fe^(3+)|Fe^(2+))^(@)=-0.7120V`
The value of cell Ecell will be -

A

`-0.3125V`

B

`0.3125V`

C

`1.114V`

D

`-1.114V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_("cell")^(@)=E_(Ag^(+)|Ag)-E_(Fe^(3+)|Fe^(2+))`
`=-0.3995+0.7120=0.3125V`
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