When a $Zn$ piece is kept in a $CuSO_4$ solution,the copper gets precipitated because the standard reduction potential of zinc is

The standard electrode potential of the cell $H_2 | H^{+} || Ag^{+} | Ag$ is .......... $V$.

Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)

Given the standard electrode potentials,$K^{+}/K = -2.93 \, V$,$Ag^{+}/Ag = 0.80 \, V$,$Hg^{2+}/Hg = 0.79 \, V$,$Mg^{2+}/Mg = -2.37 \, V$,and $Cr^{3+}/Cr = -0.74 \, V$,arrange these metals in their increasing order of reducing power.

The $E^{0}_{Red}$ values for elements $I, II, III,$ and $IV$ are $-3.04 \, V, -1.90 \, V, 0.00 \, V,$ and $1.90 \, V$ respectively. Which element has the maximum reducing power?

Given the standard half-cell potentials $(E^{\circ})$ as: $Zn \rightarrow Zn^{2+} + 2e^{-}$; $E^{\circ} = +0.76 \ V$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$; $E^{\circ} = +0.41 \ V$. Then the standard e.m.f. of the cell with the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is: