Mn${\mathrm{O}}_4^{-}$(aq) + 8H⁺(aq) + 5e^- $\to$Mn²⁺(aq) + 4H₂O(l);${\mathrm{E}}^{°}$=1051V

Cr₂O_7^2-(aq) +14H⁺(aq) + 6e _$\to$2Cr^₃₊(aq) ₊ 7H₂O(l)_{;${\mathrm{E}}^{°}$}=1.38V

Fe³⁺(aq) + e^- _$\to$Fe²⁺(aq);_{${\mathrm{E}}^{°}$}=0.77V

Cl₂(g) + 2e^- _$\to$2Cl^-(aq); _{${\mathrm{E}}^{°}$}= 1.40V

The incorrect statement regarding the quantitative estimation of aqueous  Fe(NO₃)_{2 is -}

1. Mn${\mathrm{O}}_4^{-}$ can be used in aqueous HCl

2. Cr₂O_7^2- can be used in aqueous HCl

3. MnO^-2₄ can be used in aqueous H₂SO₄

4.  Cr₂O_7^2- can be used in aqueous H₂SO₄

 The strongest oxidising agent among the following is -

1. $Br{O}_3^{-}/B{r}^{2+},E^o=+1.50$

2. $F{e}^{3+}/F{e}^{2+},E^o=+0.76$

3. $Mn{O}_4^{-}/M{n}^{2+},E^o=+1.52$

4. $C{r}_2{O}_7^{2-}/C{r}^{3+},E^o=+1.33$

 ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.35$ $\mathrm{V}$
${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-};$ ${\mathrm{E}}^{\mathrm{o}}=-0.77$ $\mathrm{V}$
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. ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}$

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:

Based on standard electrode potentials given above, the correct arrangement for increasing order of reducing power of
elements 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

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 if $F{e}^{+2},$ $F{e}^{+3}$, and Fe  block 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

The correct statement about the given reaction is-

(CN)_2(g) + 2OH^-_(aq) $\mathbf{\to }$CN^-_(aq) + CNO^-_(aq) + H₂O_(l)

The ${\mathrm{Mn}}^{3+}$ ion is unstable in solution and undergoes disproportionation reaction to give ${\mathrm{Mn}}^{2+},$ ${\mathrm{MnO}}_2$ $\mathrm{and}$ ${\mathrm{H}}^{+}$ ion. The balanced ionic equation for the reaction is-