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Consider the reaction : `H_(2)(g),Pt|H^(+)(aq) , E=0.1" V"` The pH of the solution is |
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Answer» `H_(2)(g)+2e^(-) to2H^(+)(aq)` `E=E^(@)-(0.0591)/(2)"log"[H^(+)]^(2)` `0.1=0-(0.0591)/(2)2log[H^(+)]` `0.1=0-0.0591" log " [H^(+)]` `-log[H^(+)]=(0.1)/(0.0591)=1.69` `pH=-log[H^(+)]=1.69=2` |
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