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Hydrogen electrode is placed in the solu...

Hydrogen electrode is placed in the solution whose pH is 10. The potential of this electrode will be

A

`+0. 591 V`

B

`- 0.591 V`

C

`0`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(H^(+)//1/2H_(2))=E_(H^(+)//1/2H_(2))^(@)-0.0591/1"log "1/10^(-10)`
`=-0.0591xx10=-0.591 V`
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