The standard electrode potentials of the half-cells are given below:
$Zn^{2+} + 2e^- \to Zn$; $E^{\circ} = -0.76 \, V$
$Fe^{2+} + 2e^- \to Fe$; $E^{\circ} = -0.44 \, V$
The $EMF$ of the cell $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

The standard reduction potential $E^{\circ}$ for half reactions are

$Zn \rightarrow Zn^{2+} + 2e^-$	$E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$	$E^{\circ} = +0.41 \ V$

The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

Standard electrode potential of $NHE$ at $298\,K$ is .............. $V$.

For the given reactions:
$Sn^{2+} + 2e^{-} \rightarrow Sn$
$Sn^{4+} + 4e^{-} \rightarrow Sn$
The electrode potentials are $E^{\circ}_{Sn^{2+}/Sn} = -0.140 \ V$ and $E^{\circ}_{Sn^{4+}/Sn} = 0.010 \ V$. The magnitude of standard electrode potential for $Sn^{4+}/Sn^{2+}$,i.e.,$E^{\circ}_{Sn^{4+}/Sn^{2+}}$,is $..... \times 10^{-2} \ V$. (Nearest integer)

For the reaction $M^{+n}_{(aq)} + ne^{-} \rightarrow M_{(s)}$,if the standard reduction potentials of elements $M_1, M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively,what is the order of their reducing strength?

Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$