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`Pt(H_(2))(1atm)|H_(2)O`, electrode potential at `298K` isA. `-0.2364V`B. `-0.4137V`C. `0.4137V`D. `0.00V` |
Answer» Correct Answer - c Since oxidation occurs. `H_(2) rarr 2H^(o+)+2e^(-)` `Q=[H^(o+)]^(2)=10^(-14)M^(2) i n H_(2)O` ` E=E^(c-)-(0.0591)/(2)log10^(-14)` `=0+0.0.591xx7=0.4137V` |
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