1.

The oxidation potential of a hydrogen electrode at pH=10 and pH_(1)=1

Answer»

0.059V
0.59V
0.00V
0.51V

Solution :`E_(OP)=E_(OP)^(o)-(0.059)/(1)"LOG"([H^(+)])/(P_(H_(2)))`
`therefore[H^(+)]=10^(-10),P_(H_(2))=1ATM,E_(OP)=0.59V`


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