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 $V$)?

Which one of the following has a potential more than zero?

The reduction potential of a hydrogen electrode at $25^{\circ}C$ is ............... $V$ $(P_{H_2} = 1 \ atm; [H^+] = 0.1 \ M)$.

$Pt | H_2 (1 \ atm) | H^{+} (0.001 \ M) || H^{+} (0.1 \ M) | H_2 (1 \ atm) | Pt$. What will be the value of $E_{cell}$ for this cell? ............. $V$

Consider the following electrodes $P = Zn^{2+}(0.0001 \ M) / Zn$,$Q = Zn^{2+}(0.1 \ M) / Zn$,$R = Zn^{2+}(0.01 \ M) / Zn$,$S = Zn^{2+}(0.001 \ M) / Zn$. Given $E^{\circ}(Zn^{2+} / Zn) = -0.76 \ V$,the electrode potentials of the above electrodes in volts are in the order:

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)