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The two half cell reaction of an elect...

The two half cell reaction of an electrochemical cell is given as
`Ag^(+)+e^(-)rarrAg ,E_(Ag^(+))^(@)=-0.3995 V`
`Fe^(2+)rarrFe^(3+)+e^(-),E_(Fe^(3+))^(@)//Fe^(2+)=-0.7120V`
The value of cell EMF will be

A

`-0.3125 V`

B

`0.3125 V`

C

`1.114 V`

D

`-1.114 V`

Text Solution

Verified by Experts

The correct Answer is:
B
Species with more negative `E^(@)` (standard reduction potential ) generally acts as reducing agent while with less negatvie `E^(@)` acts oxidising lagent thus the overall reaction is
`Ag^(+) +Fe^(+) rarrFe^(3+) +Ag`
`triangle E^(@) =E_(red)^(@)` =-0.3995 -(-0.7120)V`
`=-0.3995 + 0.7120 V=+0.3125 V`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-ELECTROCHEMISTRY-Exercise 1
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