Consider the change in the oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

\(\small{BrO_4^-\ \overset{1.82V}{\longrightarrow}\ BrO_3^-\ \overset{1.5V}{\longrightarrow} HBrO\ \overset{1.595V}{\longrightarrow}\ Br_2 \overset{1.0652V}{\longrightarrow}\ Br^-}\) 
Then the species undergoing disproportionation is:-

1. ${\mathrm{BrO}}_3^{-}$

2. ${\mathrm{BrO}}_4^{-}$

3. ${\mathrm{Br}}_2$

4. HBrO

Iron does not get oxidised in:

The respective oxidation states of P in H₄P₂O₅, H₄P₂O₆, and H₄P₂O₇ are: 

On the basis of the following E^O values, the strongest oxidizing agent is 
 ${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}\to {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-}; {\mathrm{E}}^{\mathrm{o}}=-0.35 \mathrm{V}$
${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}; {\mathrm{E}}^{\mathrm{o}}=-0.77 \mathrm{V}$
 

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 number of moles of KMnO₄ that will be needed to react with one mole of sulphite ion in acidic solution is :

1. $\frac{3}{5}$

2. $\frac{4}{5}$

3. $\frac{2}{5}$

4. $1$

The oxidation number of Cl in CaOCl₂
is
1. Zero

2. –1

3. +1

4. –1 as well as +1

In the conversion, H${}_2$SO${}_4$ $\to$ H${}_2$S${}_2$O${}_8$which process occurs?
1. Oxidation
2. Reduction
3. Oxidation as well as reduction
4. Neither oxidation nor reduction

Which acts as reducing agent only?

1. SO${}_2$

2. H${}_2$SO${}_4$

3. H${}_2$S

4. HNO${}_2$

In the reaction, 2 Cu${}^{+}$ $\to$ Cu + Cu${}^{2+}$, the equivalent weight of Cu${}^{+}$ is (M is the mol. wt. of Cu${}^{+}$)

1. $\mathrm{M}$

2.$\frac{\mathrm{M}}{2}$

3. $\frac{\mathrm{M}}{4}$

4. $2\mathrm{M}$