If $E^{\circ}(Fe^{+2}_{(aq)} \mid Fe_{(s)}) = -0.44 \ V$ and $E^{\circ}(Sn^{+2}_{(aq)} \mid Sn_{(s)}) = -0.14 \ V$,what is the standard $emf$ of the cell containing the two electrodes?

Given:
$E^o_{Fe^{3+} /Fe} = -0.036 \ V, E^o_{Fe^{2+} /Fe} = -0.439 \ V$
The value of standard electrode potential for the change,
$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ will be ........ $V$.

The standard electrode potentials $(E^{0}_{Red})$ of four elements $A, B, C,$ and $D$ are $-3.05 \ V, -1.66 \ V, -0.40 \ V,$ and $0.80 \ V$ respectively. Which of the following is the most reactive?

Calculate $\Delta G^{\circ}$ for the cell: $Sn_{(s)} | Sn^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ at $25^{\circ} C$ given that $E^{\circ}_{cell} = 0.90 \ V$. (in $kJ$)

Standard electrode potential values,$E^{\Theta}$ for $Al^{3+}/Al$ is $-1.66 \ V$ and that of $Tl^{3+}/Tl$ is $+1.26 \ V$. Predict the formation of $M^{3+}$ ion in solution and compare the electropositive character of the two metals.

The standard reduction potential at $298 \ K$ for the following half cells are given below :-
$NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO_{(g)} + 2H_2O \quad E^0 = 0.97 \ V$

$V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$	$E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$	$E^0 = -0.04 \ V$
$Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$	$E^0 = 0.80 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$	$E^0 = 1.40 \ V$

The number of metal$(s)$ which will be oxidized by $NO_3^{-}$ in aqueous solution is $....$.