Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

Given at $298 \ K$: $E^\ominus_{Fe^{2+}/Fe} = X \ V$; $E^\ominus_{Fe^{3+}/Fe} = Y \ V$. The $E^\ominus_{Fe^{3+}/Fe^{2+}}$ in Volt at $298 \ K$ is given by:

The $E^o$ values of half-cells are given below. Which combination of two half-cells will form a cell with the maximum potential?
$(i)$ $A + e^- \rightarrow A^- ; E^o = -0.24 \ V$
$(ii)$ $B^- + e^- \rightarrow B^{2-} ; E^o = +1.25 \ V$
$(iii)$ $C^- + 2e^- \rightarrow C^{3-} ; E^o = -1.25 \ V$
$(iv)$ $D + 2e^- \rightarrow D^{2-} ; E^o = +0.68 \ V$

Why does copper not replace hydrogen from acids?

For the reaction $M^{+n}_{(aq)} + ne^{-} \rightarrow M_{(s)}$,if the standard reduction potentials of elements $M_1, M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively,what is the order of their reducing strength?

If the standard reduction potentials of $Zn$, $Ni$, and $Fe$ are $-0.76 \text{ V}$, $-0.23 \text{ V}$, and $-0.44 \text{ V}$ respectively, determine the electrodes $X$ and $Y$ for the reaction $X(s) + Y^{+2}(aq) \rightarrow X^{+2}(aq) + Y(s)$ to be spontaneous.