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Given electrode potentials are: Fe^(3+) ...

Given electrode potentials are:
`Fe^(3+) +e rarrFe^(2+) ,E^@=0.771V`
`I_2 +2e rarr 2I^- ,E^@=0.536V`
`E^@` cell for the cell reaction,
` 2Fe^(3+) +2I^- rarr2Fe(2+) +I_2`is:

A

`(2xx0.771 0.536)=1.006V`

B

`(0.771 -0.5xx0.536)=0.503V`

C

`0.771-0.536=0.235 V`

D

`0.536-0.771=-0.236 V`

Text Solution

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The correct Answer is:
A
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