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

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1 Answers

Correct Answer - 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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