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For the half cell At pH=2. Electro...

For the half cell

At `pH=2`. Electrod potential is `:`

A

`1.36V`

B

`1.30 V`

C

`1.42 V`

D

`1.20 V`

Text Solution

Verified by Experts

The correct Answer is:
3

`E=E^(@)-(0.059)/(2)"log of "[H^(+)]^(2)`
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