The hydrogen electrode is dipped in a solution of $pH = 3$ at $25\,^oC$. The potential of the cell would be ............. $V$ (the value of $2.303\,RT/F$ is $0.059\,V$).

Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$

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: $Fe^{3+} + e^- \longrightarrow Fe^{2+}$,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771\,V$
$\frac{2.303 RT}{F} = 0.06\,V$

Consider the following cell reaction:
$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$
At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

What will be the electrode potential of a $Cu$ electrode dipped in $0.025 \ M \ CuSO_4$ solution at $298 \ K$,given that the standard reduction potential of $Cu$ is $0.34 \ V$ (in $V$)?

The hydrogen electrode is dipped in a solution of $pH=3$ at $25^{\circ} C$. The potential of the electrode will be . . . . . . $\times 10^{-2} \ V$. $\left(\frac{2.303 RT}{F}=0.059 \ V\right)$