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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 (c ) `E=E^(@)-(0.0591)/(2)log [H^(+)]^(2)` `=1.30-(0.0591)/(2)log(10^(-2))^(2)` `=E^(@)-(0.0591)/(2)(-4" log "10)` `=1.30+(0.236)/(2)=1.418" V"` |
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