Redox Reactions Test

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Redox Reactions Test - 58

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Question 1
1 / -0

For the redox reaction
$$Mn{O}_{4}^{-}+{C}_{2}{O}_{4}^{2-}+{H}^{+}\rightarrow {Mn}^{2+}+{CO}_{2}+{H}_{2}O$$ the correct coefficients of the reactants for the balanced reaction are

A
$$Mn{O}_{4}^{-}:2$$; $${C}_{2}{O}_{4}^{2-}:5$$; $${H}^{+}:16$$

B
$$Mn{O}_{4}^{-}:16$$; $${C}_{2}{O}_{4}^{2-}:5$$; $${H}^{+}:2$$

C
$$Mn{O}_{4}^{-}:5$$; $${C}_{2}{O}_{4}^{2-}:16$$; $${H}^{+}:2$$

D
$$Mn{O}_{4}^{-}:2$$; $${C}_{2}{O}_{4}^{2-}:16$$; $${H}^{+}:5$$

Solution

$$Mn{O}_{4}^{-}+{C}_{2}{O}_{4}^{2-}+{H}^{+}\rightarrow {Mn}^{2+}+{CO}_{2}+{H}_{2}O$$
$${C}^{2}{O}_{4}^{2-}\rightarrow 2{CO}_{2}+2{e}^{-}\times 5$$
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$$2Mn{O}_{4}^{-}+5{C}_{2}{O}_{4}^{2-}+16{H}^{+}\rightarrow 2{Mn}^{2+}+10{CO}_{2}+8{H}_{2}O$$
Thus the coefficient of $$Mn{O}_{4}^{-}$$, $${C}_{2}{O}_{4}^{2-}$$ and $${H}^{+}$$ in the above balanced equation respectively are $$2,5,16$$
Question 2
1 / -0

The oxidation states of the most electronegative element in the products of the reaction of $$Ba{O}_{2}$$ with dilute $${H}_{2}{SO}_{4}$$ are

A
$$0$$ and $$-1$$

B
$$-1$$ and $$-2$$

C
$$-2$$ and $$0$$

D
$$-2$$ and $$+1$$

Solution

In $${H}_{2}{O}_{2}$$ oxygen shows $$=-1$$ (peroxide) oxidation state and in $$Ba{SO}_{4}$$ oxygen shows $$=-2$$ oxidation state
Question 3
1 / -0

The atomic number of an element which shows the oxidation state of $$+3$$ is

A
$$13$$

B
$$32$$

C
$$33$$

D
$$17$$

Solution

For +3 oxidation state the outer orbital configuration has to be $$ns^2ns^1$$ and that is possible for atomic number 13.
$$Al$$ having atomic no. 13 shows $$+3$$ oxidation state
Question 4
1 / -0

Which of the following reactions involves oxidation-reduction

A
$$NaBr+HCl\rightarrow NaCl+HBr$$

B
$$HBr+Ag{NO}_{3}\rightarrow AgBr+H{NO}_{3}$$

C
$${H}_{2}+{Br}_{2}\rightarrow 2HBr$$

D
$$2NaOH+{H}_{2}{SO}_{4}\rightarrow {Na}_{2}{SO}_{4}+2{H}_{2}O$$

Solution

This is a redox reaction.
Question 5
1 / -0

The number of electrons involved in the reduction of $${Cr}_{2}{O}_{7}^{2-}$$ in acidic solution to $${Cr}^{3+}$$ is

A
0

B
2

C
3

D
5

Solution

$$Cr_2O_7^{2-} +3e^- \rightarrow Cr^{3+}$$
In this reaction three electrons are required for the reduction of $${Cr}_{2}{O}_{7}^{2-}$$ into $${Cr}^{3+}$$
Question 6
1 / -0

$${M}^{+3}$$ ion loses $$3{e}^{-}$$. Its oxidation number will be

A
$$0$$

B
$$+3$$

C
$$+6$$

D
$$-3$$

Solution

$$2\times $$No. of $${e}^{-}$$ losses $$=$$ oxidation no.
$$2\times 3{e}^{-}=+6$$
Question 7
1 / -0

In which of the compounds does hydrogen have an oxidation state of -1?

A
$$ CH_{4} $$

B
$$ NH_{3} $$

C
$$HCl$$

D
$$ CaH_{2} $$

Solution

$$ Ca+H_{2}\rightarrow CaH_2\to Ca^{2+}+2H^-$$

Hydrogen has -ve (-1) oxidation state as $$Ca$$ is more electropositive than $$H.$$
Question 8
1 / -0

In which of the following pair of compounds, oxidation number of chromium is same?

A
$$K_2CrO_4$$ and $$KCrO_2$$

B
$$KCrO_2$$ and $$Cr(CO)_6$$

C
$$K_2Cr_2O_7$$ and $$Cr(CO)_4$$

D
$$K_2Cr_2O_7$$ and $$K_2CrO_4$$

Solution

$$KCrO_2$$ and $$Cr(CO)_6$$.
Question 9
1 / -0

Oxidation number of sulphur in peroxy mono sulphuric acid $$(H_2SO_5)$$ is:

A
$$+4$$

B
$$+2$$

C
$$+6$$

D
$$-2$$

Solution

$$H_2SO_5$$

Oxidation number of $$O$$ in peroxo linkage $$=-1$$

Oxidation number of other $$O$$atoms $$=-2$$

$$+2+x-6-2=0$$ or $$x=+6$$
Question 10
1 / -0

The correct sequence of the oxidation state of underlined element is:

$$Na_2[Fe(CN)_5 NO], K_2 TaF_7, Mg_2, P_2O_7, Na_2 S_4O_6, N_3H$$

A
$$+3,+5,+5,+2.5,-\dfrac 13$$

B
$$+5,+3,+5,+3,+\dfrac 13$$

C
$$+3,+3,+5,+5,-\dfrac 13$$

D
$$+5,+5,+3,+2.5,+\dfrac 13$$

Solution

$$Na_2[Fe(CN)_5NO] : +2+x+5(-1)+0=0\Rightarrow x=+3$$

$$K_2TaF_7 : 1+x+(-7)=0\Rightarrow x=+5$$

$$Mg_2P_2O_7 :4+2x+(-14)=0\Rightarrow x=+5$$

$$Na_2S_4O_6 : +2+4x+(-12)=0$$ or $$4x=10\Rightarrow x=+2.5$$

$$N_3H : 3x+1=0\Rightarrow x=\dfrac 13$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 58

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Redox Reactions Test - 58

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SHARING IS CARING

Question 1

$$Mn{O}_{4}^{-}:2$$; $${C}_{2}{O}_{4}^{2-}:5$$; $${H}^{+}:16$$

$$Mn{O}_{4}^{-}:16$$; $${C}_{2}{O}_{4}^{2-}:5$$; $${H}^{+}:2$$

$$Mn{O}_{4}^{-}:5$$; $${C}_{2}{O}_{4}^{2-}:16$$; $${H}^{+}:2$$

$$Mn{O}_{4}^{-}:2$$; $${C}_{2}{O}_{4}^{2-}:16$$; $${H}^{+}:5$$

Solution

Question 2

$$0$$ and $$-1$$

$$-1$$ and $$-2$$

$$-2$$ and $$0$$

$$-2$$ and $$+1$$

Solution

Question 3

$$13$$

$$32$$

$$33$$

$$17$$

Solution

Question 4

$$NaBr+HCl\rightarrow NaCl+HBr$$

$$HBr+Ag{NO}_{3}\rightarrow AgBr+H{NO}_{3}$$

$${H}_{2}+{Br}_{2}\rightarrow 2HBr$$

$$2NaOH+{H}_{2}{SO}_{4}\rightarrow {Na}_{2}{SO}_{4}+2{H}_{2}O$$

Solution

Question 5

0

2

3

5

Solution

Question 6

$$0$$

$$+3$$

$$+6$$

$$-3$$

Solution

Question 7

$$ CH_{4} $$

$$ NH_{3} $$

$$HCl$$

$$ CaH_{2} $$

Solution

Question 8

$$K_2CrO_4$$ and $$KCrO_2$$

$$KCrO_2$$ and $$Cr(CO)_6$$

$$K_2Cr_2O_7$$ and $$Cr(CO)_4$$

$$K_2Cr_2O_7$$ and $$K_2CrO_4$$

Solution

Question 9

$$+4$$

$$+2$$

$$+6$$

$$-2$$

Solution

Question 10

$$+3,+5,+5,+2.5,-\dfrac 13$$

$$+5,+3,+5,+3,+\dfrac 13$$

$$+3,+3,+5,+5,-\dfrac 13$$

$$+5,+5,+3,+2.5,+\dfrac 13$$

Solution

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