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At pH = 2, E(("Quinhydrone"))^(@) = 1.30...

At `pH = 2, E_(("Quinhydrone"))^(@) = 1.30 V, E_("Quinhydrone")` will be :

A

`1.36 V`

B

`1.30 V`

C

`1.42 V`

D

`1.20 V`

Text Solution

Verified by Experts

The correct Answer is:
C
`E = E^(@) = (0.059)/(2)log[H^(+)]^(2)`
`= 1.30 - (0.059)/(2)log(10^(-2))^(2)`
`= 1.30 + (0.236)/(2) = 1.418 V`
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