The cell reaction involving the quinhydrone e | vedclass

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(C) The Nernst equation for the given electrode reaction is:$E = E^{\circ} - \frac{0.0591}{n} \log Q$For the reaction $C_6H_4(OH)_2 \rightleftharpoons

Vedclass · Accepted Answer

$1.10$

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(C) The Nernst equation for the given electrode reaction is:$E = E^{\circ} - \frac{0.0591}{n} \log Q$For the reaction $C_6H_4(OH)_2 ightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$n = 2$ and $Q = [H^+]^2$ (assuming the ratio of quinone to hydroquinone is $1:1$ in quinhydrone).$E = 1.30 - \frac{0.0591}{2} \log [H^+]^2$$E = 1.30 - 0.0591 \log [H^+]$Since $pH = -\log [H^+]$,we have $\log [H^+] = -pH = -3$.$E = 1.30 - 0.0591 	imes (3)$$E = 1.30 - 0.1773 = 1.1227 \ V$Rounding to the nearest provided option,the value is approximately $1.12 \ V$. However,using the approximation $0.06$ instead of $0.0591$:$E = 1.30 - 0.06 	imes (3) = 1.30 - 0.18 = 1.12 \ V$.Given the options,$1.12 \ V$ is closest to $1.10 \ V$.

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?

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The cell reaction involving the quinhydrone electrode is given by the following equation:$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$What will be the electrode potential at $pH = 3$ (in $V$)?