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2Ce^(4+) + Co rightarrow 2Ce^(3+) + Co^(...

`2Ce^(4+) + Co rightarrow 2Ce^(3+) + Co^(2+), E_(cell)^(@) = 1.89V`
`E_(Co^(2+)//Co)^(@) = -0.277V`. Hence `E_(Ce^(4+)//Ce^(3+))^(@)` is

A

0.805V

B

1.61V

C

`-0.805V`

D

`-1.61V`

Text Solution

Verified by Experts

The correct Answer is:
b
Half cell reactions are
`CO rarr CO^(2+)+2e^(-)`
`2Ce^(4+)+2e^(-_rarr2Ce^(3+)`
`E_("cell")^(@)=+E_(Ca^(2+))^(@)+ "-"E_(Co^(2+)//Co)^(@)`
`1.80=E_(Co^(4+) //Co^(3+))^(@)+"-"(-0.277)`
`E^(@)Ce^(4+)=1.89-0.277=1.613V`
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