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. $2\mathrm{M}$

The oxidation number of S in Na${}_2$S${}_4$O${}_6$ is
1. 2.5

2. +2 and +3

3. Zero and +5

4. +1 and +4

For the reaction,\(\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^{-}+\mathrm{MnO}_2\)  The number of moles of \(\mathrm{MnO}_4^{-} \text {and } \mathrm{MnO}_2\) respectively yielded upon the disproportionation of one mole of \(\mathrm{MnO}_4^{2-}\) are:

1. $\frac{2}{3}$and $\frac{1}{3}$

2. $\frac{1}{3}$ and $\frac{2}{3}$

3. $\frac{1}{2}$ and $\frac{1}{2}$

4. $2$ and $1$

How many moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ are required to oxidise one mole of ${\mathrm{FeSO}}_4$ in acidic medium?

1. 6

2. 1/2

3. $\frac{1}{3}$

4. $\frac{1}{6}$

${\mathrm{KMnO}}_4$ is a strong oxidising agent in acidic medium. To provide acidic medium, ${\mathrm{H}}_2{\mathrm{SO}}_4$ is used instead of HCl. This is because

1. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is stronger acid than HCl
2. HCl is oxidised by ${\mathrm{KMnO}}_4$ to ${\mathrm{Cl}}_2$
3. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is dibasic acid
4. Rate is faster in the presence of ${\mathrm{H}}_2{\mathrm{SO}}_4$