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Given E(Fe^(3+)//Fe)^(@) = -0.36 V, E(Fe...

Given `E_(Fe^(3+)//Fe)^(@) = -0.36 V, E_(Fe^(2+)//Fe)^(@)= - 0.439V`. The value of standard electrode potential for the change , `Fe^(3+)+e^(-) rarr Fe^(2+)` will be

A

`-0.072` V

B

0.385 V

C

0.770 V

D

`-0.270` V

Text Solution

Verified by Experts

The correct Answer is:
C
(i) `Fe^(3+) + 3e^(-) rarr Fe, E_(1)^(0)= -0.036V`
`DeltaG_(1)^(@) = -3xxF xx (-0.036)`
(ii) `Fe^(2+) +2e^(-) rarr Fe, E_(2)^(@) = -0.439 V`
`Delta G_(2)^(@) = -2xxFxx(-0.439)`
Required reaction :
`Fe^(3+) +e^(-) rarr Fe^(2+), DeltaG_(3) = -1xx F xx E_(3)^(@)`
Subtracting eqn. (ii) from eqn. (i) givens the required equation so that
`DeltaG_(3)^(@)= DeltaG_(1)^(@)-DeltaG_(2)^(@)`
`-1xxFxx DeltaE_(3)^(@)= -3xxFxx(-0.036)-(-2xxF xx - 0.4399)`
`E_(3)^(@)= -0.108 + 0.878 = 0.77 V`
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