Lower the electrode potential, better is the reducing power. The electrode potential increases in the order of $$\frac{{L{i^ + }}}{{Li}}\left( { - 3.05\,V} \right),\frac{{{K^ + }}}{K}\left( { - 2.93\,V} \right),$$       $$\frac{{A{l^{3 + }}}}{{Al}},\left( { - 1.66\,V} \right),\frac{{A{g^ + }}}{{Ag}}\left( { + 0.80\,V} \right)$$       and $$\frac{{A{u^{3 + }}}}{{Au}}\left( { + 1.40\,V} \right).$$    Hence, reducing power of metals will be $$Au < Ag < Al < K < {\text{ }}Li.$$

$$4P + 3KOH + 3{H_2}O \to 3K{H_2}P{O_2} + P{H_3}$$
$$O.N$$  of $$P = 0,$$  In $$K{H_2}P{O_2}$$   it is $$+1,$$  In $$P{H_3}$$  it is $$-3.$$
Hence $$P$$  is oxidised and reduced.

In neutral metal carbonyls, metals have zero oxidation state.

$$MnO_4^ - + 2{H_2}O + 3{e^ - } \to $$     $$Mn{O_2} + 4O{H^ - }$$

More is $$E_{RP}^ \circ ,$$  more is the tendency to get itself reduced or more is oxidising power.

$${F_2}$$  being most electronegative element cannot exhibit any positive oxidation state. In $$ClO_4^ - $$  chlorine is present in its highest oxidation state i.e $$+7.$$  Therefore it does not show disproportionation reaction.

$$3B{r_2} + 6CO_3^{2 - } + 3{H_2}O \to 5B{r^ - } + BrO_3^ - + 6HCO_3^ - $$
$$O.N.$$  of $$B{r_2}$$  changes from 0 to $$–1$$ and $$+5$$  hence it is reduced as well as oxidised.

$$3B{r_2} + 6CO_3^{2 - } + 3{H_2}O \to $$      $$5B{r^ - } + BrO_3^ - + 6HCO_3^ - $$
Here $$B{r_2}$$  is both reduced and oxidised.
$$\eqalign{ & B{r_2}\left( 0 \right) \to B{r^ - }\left( { - 1} \right)\left( {{\text{reduction}}} \right) \cr & B{r_2}\left( 0 \right) \to BrO_3^ - \left( { + 5} \right)\left( {{\text{oxidation}}} \right) \cr} $$

$$\mathop {Cr}\limits^{ + 3} O_2^ - < \mathop {Cl}\limits^{ + 5} O_3^ - < \mathop {Cr}\limits^{ + 6} O_4^{2 - } < \mathop {Mn}\limits^{ + 7} O_4^ - $$

In $$Ba{O_2} + {H_2}S{O_4} \to BaS{O_4} + {H_2}{O_2}$$        all atoms are present in the same $$O.S.$$  in reactants and products