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

Which of the following transformations can be carried out by using $HI$ as a reducing agent,under acidic conditions?
[Given: $I_{2(s)} \rightarrow 2I^- ; E^{\circ} = 0.54 \ V$]
$(i)$ $Cu^+ \rightarrow Cu_{(s)} ; E^{\circ} = 0.52 \ V$
$(ii)$ $Cr^{3+} \rightarrow Cr^{2+} ; E^{\circ} = -0.41 \ V$
$(iii)$ $Fe^{3+} \rightarrow Fe^{2+} ; E^{\circ} = 0.77 \ V$
$(iv)$ $Fe^{2+} \rightarrow Fe_{(s)} ; E^{\circ} = -0.44 \ V$

Fluorine is the best oxidising agent because it has

For the reaction $H_2 (1 \, bar) + 2AgCl_{(s)} \rightleftharpoons 2Ag_{(s)} + 2H^{+} (0.1 \, M) + 2Cl^{-} (0.1 \, M)$,$\Delta G^o = -48,250 \, J$ at $25 \, ^oC$. The standard emf of cell in which the given reaction takes place is ................. $V$

Calculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)

If $Cu^{+} + e^- \to Cu$ ; $E^o = X_1$ and $Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2$,then the value of $E^o$ for $Cu^{2+} + e^- \to Cu^{+}$ will be: