The reduction potential of a half cell consisting of a $Pt$ electrode immersed in $2.0 \ M \ Fe^{2+}$ and $0.02 \ M \ Fe^{3+}$ solution (in $V$) is. Given: $\left(\frac{2.303 \ RT}{F} = 0.059, E^0_{Fe^{3+} \mid Fe^{2+}} = 0.771 \ V\right)$

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

Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$

Derive the Nernst equation for the following galvanic cell: $Ni_{(s)}|Ni_{(aq)}^{2+}\parallel Ag_{(aq)}^{+}|Ag_{(s)}$

The standard electrode potential for the following reaction is $+1.33 \ V$. What is the potential at $pH = 2.0$ for the reaction: $Cr_2O_7^{2-} (aq, 1 \ M) + 14H^{+} (aq) + 6e^{-} \rightarrow 2Cr^{3+} (aq, 1 \ M) + 7H_2O (l)$?

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$