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For the half cell At pH = 3 electro...

For the half cell

At pH = 3 electrode potential is

A

`1.30V`

B

`1.20V`

C

`1.10V`

D

`1.48V`

Text Solution

Verified by Experts

The correct Answer is:
D
In this equation `E =E^(@)-(0.0591)/(2)log[H^(+)]^(2)`
`=E^(@)-0."591 log"[H^(+)]=E^(@)+0.0591"pH"`
`=1.30+0.0591xx3=1.48V`
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