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CHXI08:REDOX REACTIONS

315067 If $H_{2} S$ gas is passed through acidified solution of $K_{2} {Cr}_{2} O_{7}$ , the colour of the solution will

1 remain unchanged

2 become deep red

3 become green

4 become deep gray

Explanation:

$K_{2} {Cr}_{2} O_{7} + 4 H_{2} {SO}_{4} + 3 H_{2} S \to$
$K_{2} {SO}_{4} + {Cr}_{2} {({SO}_{4})}_{3} + 7 H_{2} O + 3 S$
Green colour appears because of ${Cr}_{2} {({SO}_{4})}_{3}$ in solution.

CHXI08:REDOX REACTIONS

315068 A bottle of $H_{2} O_{2}$ is labelled as 10 vol $H_{2} O_{2} . 112 mL$ of this solution of $H_{2} O_{2}$ is titrated against 0.04 M acidified solution of ${KMnO}_{4}$ . The volume of ${KMnO}_{4}$ in litre is

1 1

2 2

3 5

4 10

Explanation:

5.6 ' $V$ ' $H_{2} O_{2} = 1 N$
$10 \cdot^{'} V^{'} H_{2} O_{2} = \frac{10}{5.6} N$
m Eq of $H_{2} O_{2} = m Eq$ of ${MnO}_{4}^{⊖}$
$\frac{10}{5.6} \times 112 \equiv 0.04 \times 5$ ( $n$ factor $) \times V$
$V \equiv 1000 m l = 1 L$

CHXI08:REDOX REACTIONS

315069 100 mL of ozone at STP were passed through 100 mL of 10 volume $H_{2} O_{2}$ solution. What is the volume strength of $H_{2} O_{2}$ after the reaction?

1 5.6

2 11.2

3 9.5

4 2.8

Explanation:

$\begin{array}{l} O_{3} \to O_{2} + O (1) \\ H_{2} O_{2} \to H_{2} O + O (2) \\ O + O \to O_{2} (3) \\ 1 / 2 v o l 1 / 2 v o l 1 v o l \end{array}$
From Eqs. (1) and (2),
we infer that $100 {mL}_{m}$ of $O_{3}$ at STP will produce 100 mL of molecular $O_{2}$ as such and 100 mL of oxygen molecule after reaction with $H_{2} O_{2}$ .
This new volume of 100 mL of molecular oxygen after reaction with $H_{2} O_{2}$ is contributed equally by $O_{3}$ and $H_{2} O_{2}$ . Thus, 50 mL of oxygen have been contributed by $H_{2} O_{2}$ .
Again, we know
Volume of $H_{2} O_{2} \times$ Volume strength of $H_{2} O_{2}$
$= Volume of O_{2} at STP$
$∴ 100 mL$ of $^{'} 10 V^{'} H_{2} O_{2} \equiv 1000 m L$ of $O_{2}$ at STP After utilisation of 50 mL of $O_{2}$ , according to Eq. (iii), the balance $(1000 - 50) = 950 mL$ of $O_{2}$ at STP are still retainable by 100 mL of $H_{2} O_{2}$ .
Hence volume strength of $H_{2} O_{2}$ after reaction $= \frac{Volume of O_{2} at STP}{Volume of H_{2} O_{2}} = \frac{950}{100} = 9.5 V$
$∴$ Volume strength $= 9.5$

CHXI08:REDOX REACTIONS

315067 If $H_{2} S$ gas is passed through acidified solution of $K_{2} {Cr}_{2} O_{7}$ , the colour of the solution will

1 remain unchanged

2 become deep red

3 become green

4 become deep gray

Explanation:

$K_{2} {Cr}_{2} O_{7} + 4 H_{2} {SO}_{4} + 3 H_{2} S \to$
$K_{2} {SO}_{4} + {Cr}_{2} {({SO}_{4})}_{3} + 7 H_{2} O + 3 S$
Green colour appears because of ${Cr}_{2} {({SO}_{4})}_{3}$ in solution.

CHXI08:REDOX REACTIONS

315068 A bottle of $H_{2} O_{2}$ is labelled as 10 vol $H_{2} O_{2} . 112 mL$ of this solution of $H_{2} O_{2}$ is titrated against 0.04 M acidified solution of ${KMnO}_{4}$ . The volume of ${KMnO}_{4}$ in litre is

1 1

2 2

3 5

4 10

Explanation:

5.6 ' $V$ ' $H_{2} O_{2} = 1 N$
$10 \cdot^{'} V^{'} H_{2} O_{2} = \frac{10}{5.6} N$
m Eq of $H_{2} O_{2} = m Eq$ of ${MnO}_{4}^{⊖}$
$\frac{10}{5.6} \times 112 \equiv 0.04 \times 5$ ( $n$ factor $) \times V$
$V \equiv 1000 m l = 1 L$

CHXI08:REDOX REACTIONS

315069 100 mL of ozone at STP were passed through 100 mL of 10 volume $H_{2} O_{2}$ solution. What is the volume strength of $H_{2} O_{2}$ after the reaction?

1 5.6

2 11.2

3 9.5

4 2.8

Explanation:

$\begin{array}{l} O_{3} \to O_{2} + O (1) \\ H_{2} O_{2} \to H_{2} O + O (2) \\ O + O \to O_{2} (3) \\ 1 / 2 v o l 1 / 2 v o l 1 v o l \end{array}$
From Eqs. (1) and (2),
we infer that $100 {mL}_{m}$ of $O_{3}$ at STP will produce 100 mL of molecular $O_{2}$ as such and 100 mL of oxygen molecule after reaction with $H_{2} O_{2}$ .
This new volume of 100 mL of molecular oxygen after reaction with $H_{2} O_{2}$ is contributed equally by $O_{3}$ and $H_{2} O_{2}$ . Thus, 50 mL of oxygen have been contributed by $H_{2} O_{2}$ .
Again, we know
Volume of $H_{2} O_{2} \times$ Volume strength of $H_{2} O_{2}$
$= Volume of O_{2} at STP$
$∴ 100 mL$ of $^{'} 10 V^{'} H_{2} O_{2} \equiv 1000 m L$ of $O_{2}$ at STP After utilisation of 50 mL of $O_{2}$ , according to Eq. (iii), the balance $(1000 - 50) = 950 mL$ of $O_{2}$ at STP are still retainable by 100 mL of $H_{2} O_{2}$ .
Hence volume strength of $H_{2} O_{2}$ after reaction $= \frac{Volume of O_{2} at STP}{Volume of H_{2} O_{2}} = \frac{950}{100} = 9.5 V$
$∴$ Volume strength $= 9.5$

CHXI08:REDOX REACTIONS

315067 If $H_{2} S$ gas is passed through acidified solution of $K_{2} {Cr}_{2} O_{7}$ , the colour of the solution will

1 remain unchanged

2 become deep red

3 become green

4 become deep gray

Explanation:

$K_{2} {Cr}_{2} O_{7} + 4 H_{2} {SO}_{4} + 3 H_{2} S \to$
$K_{2} {SO}_{4} + {Cr}_{2} {({SO}_{4})}_{3} + 7 H_{2} O + 3 S$
Green colour appears because of ${Cr}_{2} {({SO}_{4})}_{3}$ in solution.

CHXI08:REDOX REACTIONS

315068 A bottle of $H_{2} O_{2}$ is labelled as 10 vol $H_{2} O_{2} . 112 mL$ of this solution of $H_{2} O_{2}$ is titrated against 0.04 M acidified solution of ${KMnO}_{4}$ . The volume of ${KMnO}_{4}$ in litre is

1 1

2 2

3 5

4 10

Explanation:

5.6 ' $V$ ' $H_{2} O_{2} = 1 N$
$10 \cdot^{'} V^{'} H_{2} O_{2} = \frac{10}{5.6} N$
m Eq of $H_{2} O_{2} = m Eq$ of ${MnO}_{4}^{⊖}$
$\frac{10}{5.6} \times 112 \equiv 0.04 \times 5$ ( $n$ factor $) \times V$
$V \equiv 1000 m l = 1 L$

CHXI08:REDOX REACTIONS

315069 100 mL of ozone at STP were passed through 100 mL of 10 volume $H_{2} O_{2}$ solution. What is the volume strength of $H_{2} O_{2}$ after the reaction?

1 5.6

2 11.2

3 9.5

4 2.8

Explanation:

$\begin{array}{l} O_{3} \to O_{2} + O (1) \\ H_{2} O_{2} \to H_{2} O + O (2) \\ O + O \to O_{2} (3) \\ 1 / 2 v o l 1 / 2 v o l 1 v o l \end{array}$
From Eqs. (1) and (2),
we infer that $100 {mL}_{m}$ of $O_{3}$ at STP will produce 100 mL of molecular $O_{2}$ as such and 100 mL of oxygen molecule after reaction with $H_{2} O_{2}$ .
This new volume of 100 mL of molecular oxygen after reaction with $H_{2} O_{2}$ is contributed equally by $O_{3}$ and $H_{2} O_{2}$ . Thus, 50 mL of oxygen have been contributed by $H_{2} O_{2}$ .
Again, we know
Volume of $H_{2} O_{2} \times$ Volume strength of $H_{2} O_{2}$
$= Volume of O_{2} at STP$
$∴ 100 mL$ of $^{'} 10 V^{'} H_{2} O_{2} \equiv 1000 m L$ of $O_{2}$ at STP After utilisation of 50 mL of $O_{2}$ , according to Eq. (iii), the balance $(1000 - 50) = 950 mL$ of $O_{2}$ at STP are still retainable by 100 mL of $H_{2} O_{2}$ .
Hence volume strength of $H_{2} O_{2}$ after reaction $= \frac{Volume of O_{2} at STP}{Volume of H_{2} O_{2}} = \frac{950}{100} = 9.5 V$
$∴$ Volume strength $= 9.5$

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