The two half-cell reactions of an electrochemical cell are given as: $Ag^{+} + e^{-} \rightarrow Ag$; $E^{\circ}_{Ag^{+}/Ag} = 0.7995 \ V$ and $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$; $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.7710 \ V$. The value of cell $EMF$ will be: (in $V$)

Given:
$Co^{3+} + e^- \longrightarrow Co^{2+}; E^o = 1.81 \ V$
$Pb^{4+} + 2e^- \longrightarrow Pb^{2+}; E^o = + 1.67 \ V$
$Ce^{4+} + e^- \longrightarrow Ce^{3+}; E^o = + 1.61 \ V$
$Bi^{3+} + 3e^- \longrightarrow Bi; E^o = + 0.20 \ V$
Oxidizing power of the species will increase in the order:

Identify the reaction from the following for which the standard electrode potential of the $Cu^{+2}/Cu$ electrode is $0.34 \text{ V}$ with respect to the $SHE$?

Standard reduction potentials for the following reactions are :
$Fe^{3+} + 3e^{-} \rightarrow Fe ; E^{\circ}_{1} = -0.036 \ V$
$Fe^{2+} + 2e^{-} \rightarrow Fe ; E^{\circ}_{2} = -0.44 \ V$
What will be the standard electrode potential for
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} (in $V$)?$

$E^o$ values for the half-cell reactions are given below:
$Cu^{2+} + e^- \to Cu^{+} \quad E^o = 0.15 \ V$
$Cu^{2+} + 2e^- \to Cu \quad E^o = 0.34 \ V$
What will be the $E^o$ for the half-cell reaction:
$Cu^{+} + e^- \to Cu$ ? $\dots \ V$

$MX$ 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: $\frac{2.303 RT}{F} = 0.059 \ V$]