What would be the electrode potential for the given half-cell reaction at $pH = 5$ (in $V$)?
$2 H_2O \rightarrow O_2 + 4 H^{\oplus} + 4 e^{-} ; E_{red}^{0} = 1.23 \ V$
$(R = 8.314 \ J \ mol^{-1} \ K^{-1} ; \text{Temperature} = 298 \ K ; \text{oxygen under std. atm. pressure of } 1 \ bar)$

Write a note on the relation between Gibbs free energy and cell potential for a cell reaction.

For a certain redox reaction in a galvanic cell $X(s) + Y^{2+}(aq) \rightarrow X^{2+}(aq) + Y(s)$, $E^0_{cell} = 0.0296 \text{ V}$ at $298 \text{ K}$. What is the equilibrium constant $(K_c)$ of the reaction?

Which will increase the voltage of the cell $Sn_{(s)} + 2Ag^{+}_{(aq)} \to Sn^{2+}_{(aq)} + 2Ag_{(s)}$?

Consider the cell
$Pt_{(s)} \mid H_2(g, 1\,atm) \mid H^{+}(aq, 1\,M) \parallel Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} \mid Pt_{(s)}$
When the potential of the cell is $0.712\,V$ at $298\,K$,the ratio $[Fe^{2+}] / [Fe^{3+}]$ is $.......$ (Nearest integer).
Given: $Fe^{3+} + e^- \longrightarrow Fe^{2+}$,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771\,V$
$\frac{2.303 RT}{F} = 0.06\,V$

At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)