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 $V$)?

Calculate the $emf$ of the cell in which the following reaction takes place:
$Ni_{(s)} + 2Ag^{+}(0.002 \, M) \rightarrow Ni^{2+}(0.160 \, M) + 2Ag_{(s)}$
Given that $E^{\Theta}_{(cell)} = 1.05 \, V$ (in $, V$)

For a half-cell containing a $Pt$ rod immersed in a solution of $1 \ M$ $HA$,$O_{2(g)}$ is bubbled at $1 \ atm$. The standard reduction potential for water formation is $1.23 \ V$. Given a dissociation constant,$K_a = 1 \times 10^{-4}$ for $HA$,what is $E_{\text{Half-cell}}$ at $298 \ K$ in $V$?

The electromotive force for the cell is $Cu_{(s)} | Cu_{(0.004 \ M)}^{+2} || Cu_{(0.1 \ M)}^{+2} | Cu_{(s)}$

$Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$
The Gibbs free energy change for the above reaction at $298 \ K$ is $x \times 10^{-1} \ kJ \ mol^{-1}$;
The value of $x$ is ..... [nearest integer] $\left[\text{Given}: E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \ V; E^{\ominus}_{Sn^{2+}/Sn} = -0.14 \ V; F = 96500 \ C \ mol^{-1}\right]$

The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is