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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
(c ) `E=E^(@)-(0.0591)/(2)log [H^(+)]^(2)`
`=1.30-(0.0591)/(2)log(10^(-2))^(2)`
`=E^(@)-(0.0591)/(2)(-4" log "10)`
`=1.30+(0.236)/(2)=1.418" V"`
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