Calculate $E^o_{cell}$ for the following cell in $V$:
$Zn_{(s)} | Zn^{2+}_{(aq.)} || Ag^{+}_{(aq.)} | Ag_{(s)}$
Given: $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ ; $E^o_{Ag^{+}/Ag} = 0.80 \ V$

The standard reduction potentials at $298 \ K$ for the following half reactions are given against each:
$Zn^{2+}(aq.) + 2e^- \rightleftharpoons Zn_{(s)}$; $E^\circ = -0.762 \ V$
$Cr^{3+}(aq.) + 3e^- \rightleftharpoons Cr_{(s)}$; $E^\circ = -0.740 \ V$
$2H^{+}(aq.) + 2e^- \rightleftharpoons H_{2(g)}$; $E^\circ = 0.00 \ V$
$Fe^{3+}(aq.) + e^- \rightleftharpoons Fe^{2+}(aq.)$; $E^\circ = 0.770 \ V$
Which is the strongest reducing agent?

Standard electrode potential of three metals $X$,$Y$ and $Z$ are $-1.2 \, V$,$+0.5 \, V$ and $-3.0 \, V$ respectively. The reducing power of these metals will be

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E_{cell}^o$ is $1.89 \, V$ and $E_{Co/Co^{2+}}^o = +0.28 \, V$. If $E_{Ce^{4+}/Ce^{3+}}^o = x$,then $x$ is ............. $V$.

Arrange the following metals in the order of their decreasing standard electrode potential: $Mg, K, Ba, Ca$.

The $E^{\circ}$ of $Ce^{4+} / Ce^{3+} = 1.6 \ V$ and $Fe^{3+} / Fe^{2+} = 0.77 \ V$. The $E^{\circ}$ of the reaction where $Fe^{3+}$ oxidises $Ce^{3+}$ is: