Redox Reactions Test

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Redox Reactions Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation number of oxygen in $$K{ O }_{ 3 },{ Na }_{ 2 }{ O }_{ 2 }$$ is:

A
$$3, 2$$

B
$$1, 0$$

C
$$0, 1$$

D
$$-0.33, -1$$

Solution

$$K{ O }_{ 3 }$$
Suppose oxidation number of $$O=x,\ \Rightarrow 1+3x=0$$
$$3x=-1$$
$$x=-\dfrac { 1 }{ 3 } =-0.33$$

$${ Na }_{ 2 }{ O }_{ 2 }$$
Suppose oxidation number of $$O=x$$
$$2\times 1+2x=0$$
$$2+2x=0$$
$$2x=-2\Rightarrow x=-\dfrac { 2 }{ 2 } $$
$$x=-1$$
Question 2
1 / -0

In view of the signs of $$\triangle_{r}G^{\circ}$$ for the following reactions
$$PbO_{2} + Pb \rightarrow 2PbO, \triangle_{r}G^{\circ} < 0$$,
$$SnO_{2} + Sn \rightarrow 2SnO, \triangle G^{\circ} > 0$$
Which oxidation states are more characteristic for lead and tin?

A
For lead $$+4$$, For tin $$+2$$

B
For lead $$+2$$, For tin $$+2$$

C
For lead $$+4$$, For tin $$+4$$

D
For lead $$+2$$, For tin $$+4$$

Solution

$$\begin{array} { *{ 20 }{ c } }{ Pb{ O_{ 2 } } } \\ { +4 } \end{array}+\begin{array} { *{ 20 }{ c } }{ Pb } \\ 0 \end{array}\begin{array} { *{ 20 }{ c } }\to \\ { } \end{array}\begin{array} { *{ 20 }{ c } }{ 2PbO } \\ { +2 } \end{array}\Delta G>0\, \, { { Spontaneous } }\, \, +2\, \, is\, \, more\, \, stable$$
$$\begin{array} { *{ 20 }{ c } }{ Sn{ O_{ 2 } } } \\ { +4 } \end{array}+\begin{array} { *{ 20 }{ c } }{ Sn } \\ 0 \end{array}\begin{array} { *{ 20 }{ c } }\to \\ { } \end{array}\begin{array} { *{ 20 }{ c } }{ 2SnO } \\ { +2 } \end{array}\Delta G>0\, \, Non\, \, { { Spontaneous } }\, \, +2\, \, less\, \, stable$$
So for lead $$+2$$ and for $$Sn$$ $$+4$$
Option D is correct.
Question 3
1 / -0

A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number?

A
Cl

B
C

C
S

D
H

Solution
Question 4
1 / -0

In the conversion of $$Br_2$$ to $$Br{ O }^{-}_{3}$$ the oxidation state of bromine changes from:

A
$$0$$ to $$+5$$

B
$$-1$$ to $$+5$$

C
$$0$$ to $$-3$$

D
$$+2$$ to $$+5$$

Solution

Oxidation state of an atom refers to the total number of electrons gain or lost by the atom form a molecule.

$$Br_2$$is converted to$$BrO^-_3$$.
Oxidation state of Bromine in $$Br_2=0$$
Oxidation state of Bromine in$$BrO^-_3=+5$$

Oxidation state of bromine changes from 0 to +5.
Question 5
1 / -0

Consider the following half-cell reactions and associated standard half-cell potentials and determine the maximum voltage that can be obtained by combination resulting in spontaneous processes:
$$AuBr_{4}^{-}(aq) + 3e^{-} \rightarrow Au(s) + 4Br^{-}(aq); E^{\circ} = -0.86\ V$$
$$Eu^{3+}(aq) + e^{-} \rightarrow Eu^{2+}(aq); E^{\circ} = -0.43\ V$$
$$Sn^{2+}(aq) + 2e^{-} \rightarrow Sn(s); E^{\circ} = -0.14\ V$$
$$IO^{-}(aq) + H_{2}O(l) + 2e^{-}\rightarrow I^{-}(aq) + 2OH^{-}(aq); E^{\circ} = + 0.49\ V$$

A
$$+ 0.72$$

B
$$+ 1.54$$

C
$$+1.00$$

D
$$+ 1.35$$

Solution

$$\begin{array}{l} AuBr_{ 4 }^{ - }\left( { aq } \right) +3{ e^{ - } }\to Au\left( s \right) +4B{ r^{ - } }\left( { aq } \right) ;\, \, E^{ \circ }=-0.86V\to Anode \\ I{ O^{ - } }\left( { ar } \right) +{ H_{ 2 } }O\left( l \right) +2{ e^{ - } }\to { I^{ - } }\left( { aq } \right) +2O{ H^{ - } }\left( { aq } \right) ;E^{ \circ }=+0.49\to cathode \\ E{ ^{ \circ }_{ cell } }=E{ ^{ \circ }_{ cathode } }-E{ ^{ \circ }_{ anode } }\Rightarrow 0.49-\left( { -0.86 } \right) \Rightarrow +1.35V \end{array}$$
Question 6
1 / -0

What is the oxidation number of tungsten in the ion $$W_6O_6Cl^{2-}_{12}$$ ?

A
2.7

B
3.3

C
3.7

D
4.3

Solution

$$W_{6}O_{6}Cl_{12}^{2-}$$

$$6\times (x)+(-2)\times 6+12(-1)=-2$$

$$6x-12-12=-2$$

$$6x-24=-2$$

$$6x=24-2$$

$$6x=22$$

$$x=\dfrac{22}{6}$$= $$3.666$$

$$x=3.67$$

$$x=3.7$$
Question 7
1 / -0

$$CH_3COOH$$ is neutralized by $$NaOH$$. Conductometric titration curve will be of the type:

A

B

C

D

Solution

When $$CH_3COOH$$ is neutralized with $$NaOH$$. The conductometric curve is as follows-
This is because, weak acid dissouates partiaaly into $$H^+$$ & $$CH_3COO^-$$ ions & show no, of ions few in the solution. Hence the solution has low conductance value. When, few drops of strong base which dissociates as $$(Na^+ + OH^-)$$ completely acid-base reaction occurs, the $$OH^-$$ ion consumes $$H^+$$ to form $$H_2O$$. So, by adding few drops of base at first conductance decrease slightly, ith further addition of more drops of base all $$H^+$$ are consumed & more drops of base, all $$H^+$$ are consumed & dissociation of weak acid increases which is reflected by gradual increase in conductance.
Question 8
1 / -0

The oxidation numbers of the sulphur atoms in peroxomonosulphuric acid $$(H_{2}SO_{5})$$ and peroxodisulphuric acid $$(H_{2}S_{2}O_{8})$$ are respectively

A
$$+8$$ and $$+7$$

B
$$+3$$ and $$+3$$

C
$$+6$$ and $$+6$$

D
$$+4$$ and $$+6$$

Solution

Both the peroxomonosulphuric acid $$(H_{2}SO_{5})$$ and peroxodisulphuric acid $$(H_{2}S_{2}O_{8})$$ have peroxide linkage and have the following structures.

Thus, the oxidation number of $$S$$ in $$H_{2}SO_{5}$$ is calculated as:

$$\underset {(for\ H)}{2\times (+1)} + \underset {(for\ S)}{X} + \underset {for\ O}{3\times (-2)} + \underset {(for\ peroxide)}{2\times (-1)}=0$$

$$2 + x - 6 - 2 = 0$$
$$\therefore x = +6$$

Thus, oxidation number of $$S$$ in $$H_{2}S_{2}O_{8}$$ is calculated as

$$\underset {(for\ H)}{2\times (+1)} + \underset {(for\ S)}{2\times (x)} + \underset {(for\ O)}{6\times (-2)} + \underset {(for\ peroxide)}{2\times (-1)} = 0$$

$$2 + 2x - 12 + 2 = 0$$

$$\therefore x = + 6$$.
Question 9
1 / -0

Oxidation numbers of $$Mn$$ in its compounds $$ MnCl_2$$ , $$ Mn(OH)_3$$, $$MnO_2$$ and $$ KMnO_4 $$ respectively are:

A
+2, +4 ,+7 , +3

B
+2, +3, +4, +7

C
+7, +3, +2, +4

D
+7, +4, +3, +2

Solution

Consider the oxidation state of $$Mn$$ as $$x$$.
The charges on $$Cl, O, H, K$$ are $$-1, -2, +1, +1$$ respectively. Therefore, the oxidation states of $$Mn$$ in its given compounds can be calculated as follows:

$$MnCl_2$$ $$\rightarrow$$ $$x + (-2) = 0 $$ $$\Rightarrow$$ $$ x = +2$$
$$Mn(OH)_3 \rightarrow x + (-3) = 0 \Rightarrow x = +3$$
$$MnO_2 \rightarrow x + (-4) = 0 \Rightarrow x = +4$$
$$KMnO_4 \rightarrow +1 + x + (-8) = 0 \Rightarrow x = +7$$
Hence, option B is the correct answer of this question.
Question 10
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The oxidation state of sulphur in the anions $$ { SO }_{ 3 }^{ 2- },{ SO }_{ 4 }^{ 2- },S_2{ O }_{ 4 }^{ 2- },S_2{ O }_{ 6 }^{ 2- } $$ is in the order :

A
$$ S_2{ O }_{ 4 }^{ 2- } > S_2{ O }_{ 6 }^{ 2- } >{ SO }_{ 4 }^{ 2- } > { SO }_{ 3 }^{ 2- } $$

B
$$ S_2{ O }_{ 6 }^{ 2- } > { SO }_{ 3 }^{ 2- } >S_2{ O }_{ 4 }^{ 2- } > { SO }_{ 4 }^{ 2- } $$

C
$$ { SO }_{ 4 }^{ 2- } > S_2{ O }_{ 6 }^{ 2- } > { SO }_{ 3 }^{ 2- } > S_2{ O }_{ 4 }^{ 2- } $$

D
$$ { SO }_{ 3 }^{ 2- } > { SO }_{ 4 }^{ 2- } > S_2{ O }_{ 4 }^{ 2- } > S_2{ O }_{ 6 }^{ 2- } $$

Solution

Consider the oxidation state of sulphur as $$x$$. We know that the charge on oxygen atom is $$-2$$. Hence the oxidation states of sulphur in the given four anions are calculated as follows:
$$S{ O }_{ 3 }^{ 2- } \rightarrow x + (-6) = -2 \Rightarrow x = +4$$
$$S{ O }_{ 4 }^{ 2- } \rightarrow x + (-8) = -2 \Rightarrow x = +6$$
$$S_2{ O }_{ 4 }^{ 2- } \rightarrow 2x + (-8) = -2\Rightarrow x = +3$$
$$S_2{ O }_{ 6 }^{ 2- } \rightarrow 2x + (-12) = -2 \Rightarrow x = +5$$
Hence $$S{ O }_{ 4}^{ 2- } > S_2{ O }_{ 6 }^{ 2- } > S{ O }_{ 3 }^{ 2- } > S_2{ O }_{ 4 }^{ 2- }$$