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Given electrode potentials asre Fe^(3+...

Given electrode potentials asre
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.771V`
I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.536V`
`E^(c-)._(cell)` for the cell reaction,
`2Fe^(3+)+2I^(c-) rarr Fe^(2+)+I_(2)` is

A

`(2 xx 0.771 - 0.536 ) = 1.006` V

B

`(0.771 - 0.5 xx 0.536) = 0.503 V`

C

`0.771 - 0.536 = 0.235 V `

D

`0.536 - 0.771 = -0.236 V `

Text Solution

Verified by Experts

The correct Answer is:
C
`E_(cell)^(@) = E_(RHS)^(@) - E_(LHS)^(@) = E_("Higher" RP)^(@) - E_("Lower" RP)^(@)`
`=0.771 - 0.536 = +0.235` V
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