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For the half cell At pH=2, the elec...

For the half cell

At `pH=2`, the electrode potential is

A

`1.36V`

B

`1.30V`

C

`1.42V`

D

`1.20V`

Text Solution

Verified by Experts

The correct Answer is:
c
In this electrode,
`[A]=[B]` in quinhydrone electrode
Hence, `Q=[H^(o+)]^(2)`

`E=E^(c-)-(0.0591)/(2)log[H^(o+)]^(2)`
`=E^(c-)+0.0591log [H^(o+)]`
`=E^(c-)+0.0591pH`
`1.30+0.059xx2`
`=1.42V`
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