For the reaction taking place in the cell:
$Pt_{(s)} | H_{2(g)} | H^{+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$
$E_{Cell}^0 = +0.5332 \ V$.
The value of $\Delta_f G^0$ in $kJ \ mol^{-1}$ (nearest integer) is:

Given: $E^{0}_{Mn^{+7} \mid Mn^{+2}} = 1.5 \ V$ and $E^{0}_{Mn^{+4} \mid Mn^{+2}} = 1.2 \ V$,then $E^{0}_{Mn^{+7} \mid Mn^{+4}}$ is (in $V$)

$A$ standard hydrogen electrode has zero electrode potential because

$Cu^{+} + e^- \to Cu$ ; $E^o = X_1 \ V$
$Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2 \ V$
Then for $Cu^{2+} + e^- \to Cu^{+}$ ; $E^o$ will be ?

The standard electrode potentials of $Ag^{+}/Ag$,$Hg_2^{2+}/2Hg$,$Cu^{2+}/Cu$,and $Mg^{2+}/Mg$ are $0.80 \ V$,$0.79 \ V$,$0.34 \ V$,and $-2.37 \ V$,respectively. An aqueous solution containing $1 \ M$ concentration of each of these metal salts is electrolyzed. With increasing voltage,what is the correct sequence of deposition of the metals at the cathode?

The $E^{\circ}_{cell}$ of $Cu_{(s)} | Cu^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ is $0.647 \ V$. Calculate the $E^{\circ}_{Ag}$ if $E^{\circ}_{Cu}$ is $0.153 \ V$. (in $V$)