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At `pH = 2, E_(("Quinhydrone"))^(@) = 1.30 V, E_("Quinhydrone")` will be : A. `1.36 V`B. `1.30 V`C. `1.42 V`D. `1.20 V` |
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Answer» Correct Answer - C `E = E^(@) = (0.059)/(2)log[H^(+)]^(2)` `= 1.30 - (0.059)/(2)log(10^(-2))^(2)` `= 1.30 + (0.236)/(2) = 1.418 V` |
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