Calculate the standard cell potential (in $V$) of the cell in which the following reaction takes place:
$Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \to Fe^{3+}_{(aq)} + Ag_{(s)}$
Given that:
$E^o_{Ag^{+}/Ag} = x \ V$
$E^o_{Fe^{2+}/Fe} = y \ V$
$E^o_{Fe^{3+}/Fe} = z \ V$

Given the electrode potentials:
$Fe^{3+} + e^- \to Fe^{2+}; E^o = 0.771 \ V$
$I_2 + 2e^- \to 2I^{-}; E^o = 0.536 \ V$
The $E^o_{\text{cell}}$ for the cell reaction $2Fe^{3+} + 2I^{-} \to 2Fe^{2+} + I_2$ is:

The $E^o$ for half cells $Fe/Fe^{2+}$ and $Cu/Cu^{2+}$ are $-0.44 \, V$ and $+0.32 \, V$ respectively. Then

The e.m.f. of a cell whose half cells are given below is .............. $V$
$Mg^{2+} + 2e^- \to Mg_{(s)}$; $E^o = - 2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}$; $E^o = + 0.34 \ V$

$MX$ is a sparingly soluble salt that follows the given solubility equilibrium at $298 \ K$: $MX_{(s)} \rightleftharpoons M^{+}_{(aq)} + X^{-}_{(aq)}$; $K_{sp} = 10^{-10}$. If the standard reduction potential for $M^{+}_{(aq)} + e^- \rightarrow M_{(s)}$ is $(E^{\ominus}_{M^{+}/M}) = 0.79 \ V$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\ominus}_{X^{-}/MX_{(s)}/M}$ is . . . . . . $mV$. (nearest integer) [Given: $\frac{2.303 RT}{F} = 0.059 \ V$]

Which one among the following is the strongest reducing agent?
$Fe^{2+} + 2e^- \to Fe \ (E^\circ = -0.44 \ V)$
$Ni^{2+} + 2e^- \to Ni \ (E^\circ = -0.25 \ V)$
$Sn^{2+} + 2e^- \to Sn \ (E^\circ = -0.14 \ V)$
$Fe^{3+} + e^- \to Fe^{2+} \ (E^\circ = +0.77 \ V)$