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The potential of a hydrogen electrode at pH=10 is :A. `+0.59" V "`B. `0.00" V"`C. `-0.59" V "`D. `-0.059" V "` |
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Answer» Correct Answer - C (c ) `H^(+)+e^(-) to 1//2 H_(2)` `E=E^(@)-(0.059)/(n)"log"(1)/([H^(+)])=-(0.059)/(1)pH` `=(0-0.059)/(1)xx10=0-0.59=-0.59" V"` |
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