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For the half-cell : At pH= 2, electrode...

For the half-cell :
At pH= 2, electrode oxidation potential is :

A

`(1.36)`

B

`(1.30)`

C

`(1.42)`

D

`(1.2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`2H^(+)+ 2e^(-)E^(@) = 1.3V`
`E_("Oxidation")= E_("Oxidation")^(circ)-(0.0591)/(2) log_(10) ([H^(+)]^(2) = 1.3 + 0.059 PH= 1.3 + 0.059xx2 = 1.42V`
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