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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

`-1.64 V`

B

`+1.64 V`

C

`-2.08 V`

D

`+2.17 V`

Text Solution

Verified by Experts

The correct Answer is:
B
`Ce^(4+) + e^(-) to Ce^(3+)`
`Co to Co^(2+) + 2e^(-)`
`2 xx` eq (i) + eq. (ii)
`2Ce^(4+) + Co to 2Ce^(3+) + Co^(2+)`
`E_("cell")^(@) = E_("cathode"^(@) - E_("anode")^(@)`
`1.89 V =E_(Ce^(4+)//Ce^(3+)) - E_(Co^(2+)//Co)^(@)`
= 1.64 V
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