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The pressure of `H_(2)` required to make the potential of `H_(2)` -electrode zero in pure water at `50^(@)`C is :-A. `10^(-14)`B. `10^(-7)`C. `10^(+14)`D. `10^(+7)` |
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Answer» Correct Answer - A `2H^(+)(aq)+2e^(-)toH_(2)(g)` `E=E^(@)-(2.303xx8.314xx323)/(2xx96500)"log"(P_(H2))/([H^(+)]^(2))` `O=O-(2.303xx8.314xx323)/(2xx96500)"log"(P_(H2))/([H^(+)]^(2))` `(P_(H2))/10^(-14)=1` `pH_(2)=10^(-14)atm` |
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