The standard electrode potential (E°) values of  Al³⁺/ Al,  Ag⁺ / Ag, K⁺ / K,  and Cr³⁺ / Cr are –1.66 V, 0.80 V, –2.93 V, & –0.79 V respectively. The correct decreasing order of the reducing power of the metal is:

Standard reduction potentials of the half-reactions are given below: 

$F_{2\left(g\right)}$ $+$ $2e^{-}\to 2{F^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+2.85$ $V$ $$

$C{l}_{2\left(g\right)}$ $+$ $2e^{-}\to 2{C{l}^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+1.36$ $V$ $$

$B{r}_{2\left(g\right)}$ $+$ $2e^{-}\to 2{B{r}^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+1.06$ $V$ $$

$I_{2\left(g\right)}$ $+$ $e^{-}\to 2{I^{-}}_{\left(aq\right)}$ $;$ $E°$ $=+0.53$ $V$ $$

The strongest oxidizing and reducing agents, respectively, are: 

1. $B{r}_2$ and $C{l}^{-}$

2. $C{l}_2$ and $B{r}^{-}$

3. $C{l}_2$ and $I_2$

4. $F_2$ and $l^{-}$

A solution contains Fe²⁺, Fe³⁺ and I^– ions. This solution was treated with iodine at 35^°C. E° for Fe³⁺/Fe²⁺ is +0.77 V and E° for I₂/2I^– = 0.536 V.
The favourable redox reaction is:

1. Fe²⁺ will be oxidized to Fe³⁺.

2. I₂ will be reduced to I^–.

3. There will be no redox reaction.

4. I^– will be oxidized to I₂.

 ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.35$ $\mathrm{V}$
${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.77$ $\mathrm{V}$
The strongest oxidizing agent in the above equation is:

1. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}$

2. ${\mathrm{Fe}}^{2+}$

3. ${\mathrm{Fe}}^{3+}$

4. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

Standard electrode potentials are:

$F{e}^{+2}/Fe$ ,$E°$ $=$ $-0.\mathrm{44 }$

$\mathrm{ }F{e}^{+3}/F{e}^{+2},E°$ $=0.77$

Choose the correct observation when $F{e}^{+2},$ $F{e}^{+3}$, and Fe _(solid) are kept together:

1. $F{e}^{+3}$ increases 

2. $F{e}^{+3}$ decreases 

3. $\frac{F{e}^{+2}}{F{e}^{+3}}$ remains unchanged 

4. $F{e}^{+2}$ decreases