The half-cell reaction involving the quinhydr | vedclass

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Question

(A) The oxidation half-cell reaction is: $Hydroquinone \rightleftharpoons Quinone + 2H^+ + 2e^-$Using the Nernst equation for the oxidation potential

Vedclass · Accepted Answer

$1.48$

Vedclass · Answer

(A) The oxidation half-cell reaction is: $Hydroquinone ightleftharpoons Quinone + 2H^+ + 2e^-$Using the Nernst equation for the oxidation potential $(E_{op})$:$E_{op} = E_{op}^o - \frac{0.059}{n} \log_{10} Q$Here,$n = 2$ and $Q = [H^+]^2$ (assuming the ratio of quinone to hydroquinone is $1$ in quinhydrone electrode).$E_{op} = E_{op}^o - \frac{0.059}{2} \log_{10} [H^+]^2$$E_{op} = E_{op}^o - 0.059 \log_{10} [H^+]$Since $pH = -\log_{10} [H^+]$,we have:$E_{op} = E_{op}^o + 0.059 	imes pH$Given $E_{op}^o = 1.30 \ V$ and $pH = 3$:$E_{op} = 1.30 + 0.059 	imes 3$$E_{op} = 1.30 + 0.177$$E_{op} = 1.477 \ V \approx 1.48 \ V$

The half-cell reaction involving the quinhydrone electrode is shown below:
$HO-C_6H_4-OH_{(aq)} \rightleftharpoons O=C_6H_4=O_{(aq)} + 2H^+ + 2e^-$
If $E_{op}^o$ for this electrode is $1.30 \ V$,then what will be the oxidation electrode potential at $pH = 3$? ............ $V$

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The half-cell reaction involving the quinhydrone electrode is shown below:$HO-C_6H_4-OH_{(aq)} \rightleftharpoons O=C_6H_4=O_{(aq)} + 2H^+ + 2e^-$If $E_{op}^o$ for this electrode is $1.30 \ V$,then what will be the oxidation electrode potential at $pH = 3$? ............ $V$