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 $pH = 3$? ............ $V$

  • A
    $1.48$
  • B
    $1.20$
  • C
    $1.10$
  • D
    $1.05$

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