Redox Reactions Test

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Redox Reactions Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

The correct statement for the molecule $$ CsI_{3}$$ is that:

A
it contains $$ Cs^{3+}$$ and $$ I^{-}$$ ions

B
it contains $$ Cs^{+}, I^{-}$$ and lattice $$ I_{2}$$ molecule

C
it is a covalent molecule

D
it contains $$ Cs^{+}$$ and $$ I_{3}{^-}$$ ions

Solution

$$ CsI_{3} \rightarrow Cs^{+}+{I_3}^{-}$$

$$ \Rightarrow Cs$$ cannot show $$+3$$ oxidation state.

$$ \Rightarrow I_{2}$$ molecules are too large to be accommodated in the lattice.

Hence, the correct option is $$D$$
Question 2
1 / -0

Excess of KI reacts with $$CuSO_4$$ solution and then $$Na_2S_2O_3$$ solution is added to it. Which of the statement is incorrect for this reaction?

A
$$Cu_2I_2$$ is formed

B
$$CuI_2$$ is formed

C
$$Na_2S_2O_3$$ is oxidised

D
Evolved $$I_2$$ is reduced

Solution

Copper sulphate reacts with potassium iodide to form cuprous iodide and iodine.
$$\displaystyle 2CuSO_4 + 4KI \rightarrow Cu_2I_2 \downarrow + I_2 +2K_2SO_4$$
Thus, $$\displaystyle CuI_2$$ is not formed in this reaction.
Hence, the option B is incorrect and option A is correct.
The liberated iodine is titrated with sodium thiosulphate to form sodium tetrathionate.
$$\displaystyle 2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 +2NaI$$
Iodine is reduced and sodium thiosulphate is oxidized.
Question 3
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The pair of compounds having metals in their highest oxidation state is:

A
$$[Fe(CN)_6]^{3-}$$ and $$[Cu(CN)_4]^{2-}$$

B
$$MnO_2$$ and $$CrO_2Cl_2$$

C
$$[FeCl_4]^-$$ and $$Co_2O_3$$

D
$$[NiCl_4]^{2-}$$ and $$[CoCl_4]^{2-}$$
Question 4
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Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is:

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

$$\mathrm{C}\mathrm{r}_{2}\mathrm{O}_{7}^{2-}+$$ $$Fe$$ $$^{ 2+}\rightarrow$$ $$Fe$$ $$^{3+}+ \mathrm{Cr}$$ $$^{3+}$$

$$\mathrm{n}$$ factor of $$C\mathrm{r}_{2}\mathrm{O}_{7}^{2-}=6$$

$$\mathrm{n}$$ factor of $$\mathrm{F}\mathrm{e}^{2+}=1$$

So, to reduce one mole of dichromate, $$6$$ moles of $$\mathrm{F}\mathrm{e}^{2+}$$ are required.

Hence, the correct option is $$D$$
Question 5
1 / -0

The pair of compounds having metals in their highest oxidation state is:

A
$$\mathrm{M}\mathrm{n}\mathrm{O}_{2},\ \mathrm{F}\mathrm{e}\mathrm{C}1_{3}$$

B
$$[\mathrm{M}\mathrm{n}\mathrm{O}_{4}]^{-},\ \mathrm{C}\mathrm{r}\mathrm{O}_{2}\mathrm{C}1_{2}$$

C
$$[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})_{6}]^{3-},\ [\mathrm{C}\mathrm{o}(\mathrm{C}\mathrm{N})_{3}]$$

D
$$[\mathrm{N}\mathrm{i}\mathrm{C}1_{4}]^{2-},\ [\mathrm{C}\mathrm{o}\mathrm{C}1_{4}]^{-}$$

Solution

(A) In $$\mathrm{M}\mathrm{n}\mathrm{O}_{2}$$, $$Mn$$ is in +4 oxidation state.
In $$\mathrm{F}\mathrm{e}\mathrm{C}1_{3}$$, $$Fe$$ is in +3 oxidation state.
(C) In $$[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})_{6}]^{3-}$$, $$Fe$$ is in +3 oxidation state.
In $$[\mathrm{C}\mathrm{o}(\mathrm{C}\mathrm{N})_{3}]$$, $$Co$$ is in +3 oxidation state.
(D) In $$[\mathrm{N}\mathrm{i}\mathrm{C}1_{4}]^{2-}$$, $$Ni$$ is in +2 oxidation state.
In $$ [\mathrm{C}\mathrm{o}\mathrm{C}1_{4}]^{-}$$, $$Co$$ is in +3 oxidation state.
Question 6
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Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

A
$$HNO_3, NO, NH_4Cl, N_2 $$

B
$$HNO_3, NO, N_2, NH_4Cl$$

C
$$HNO_3, NH_4Cl, NO, N_2$$

D
$$NO, HNO_3, NH_4Cl, N_2$$

Solution

The oxidation states of nitrogen are given below:

$$HNO_3\rightarrow N^{+5}$$
$$NO\rightarrow N^{+2}$$
$$NH_4Cl\rightarrow N^{-3}$$
$$N_2\rightarrow N^o$$

Hence, the descending order of the oxidation state of nitrogen is $$HNO_3, NO, N_2, NH_4Cl$$.
Question 7
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$$(\mathrm{N}\mathrm{H}_{4})_{2}\mathrm{C}\mathrm{r}_{2}\mathrm{O}_{7}$$ on heating gives a gas which is also given by:

A
heating $$\mathrm{N}\mathrm{H}_{4}\mathrm{N}\mathrm{O}_{2}$$

B
heating $$\mathrm{N}\mathrm{H}_{4}\mathrm{N}\mathrm{O}_{3}$$

C
$$Mg_{3}\mathrm{N}_{2}+\mathrm{H}_{2}\mathrm{O}$$

D
$$\mathrm{N}\mathrm{a}(\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}.)+\mathrm{H}_{2}\mathrm{O}_{2}$$

Solution

$$(\mathrm{N}\mathrm{H}_{4}){}_{2}\mathrm{C}\mathrm{r}_{2}\mathrm{O}_{7}\xrightarrow[]{\Delta }\mathrm{N}_{2}\uparrow+\mathrm{C}\mathrm{r}_{2}\mathrm{O}_{3}+4\mathrm{H}_{2}\mathrm{O}$$
$$\mathrm{N}\mathrm{H}_{4}\mathrm{N}\mathrm{O}_{2}\xrightarrow[]{\Delta }\mathrm{N}_{2}\uparrow+2\mathrm{H}_{2}\mathrm{O}$$
Therefore, the answer is ammonium nitrate.
Question 8
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For the redox reaction, $$MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O$$, the correct coefficients of the reactants for the balanced equation are _____________.

A
$$MnO_{4}^{-} = 2, C_{2}O_{4}^{2-} = 16, H^{+} = 5$$

B
$$MnO_{4}^{-} = 16, C_{2}O_{4}^{2-} = 5, H^{+} = 2$$

C
$$MnO_{4}^{-} = 5, C_{2}O_{4}^{2-} = 16, H^{+} = 2$$

D
$$MnO_{4}^{-} = 2, C_{2}O_{4}^{2-} = 5, H^{+} = 16$$

Solution

The unbalanced redox reaction is $$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O$$
Balance C atoms.

$$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + 2CO_{2} + H_{2}O$$

The oxidation number of Mn decreases from $$\displaystyle +7$$ to $$\displaystyle +2$$.
The decrease in the oxidation number is $$\displaystyle 7-2=5$$
The oxidation number of C increases from $$\displaystyle +3$$ to $$\displaystyle +4$$.
The increase in the oxidation number for 1 C atom is $$\displaystyle 4-3=1$$
The increase in the oxidation number for 2 C atom is $$\displaystyle 2 \times 1=2$$

To balance the increase in the oxidation number with decrease in the oxidation number, multiply Mn containing species with 2 and C containing species with 5.

$$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} $$

Balance O atoms by adding 8 water molecules to the products side.

$$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$$

Balance H atoms by adding 16 $$\displaystyle H^+$$ ions to the reactants side.

$$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+}\rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$$

This is the balanced equation.
Question 9
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Identify Z in the reaction given below:

$$2Mn{ O}^-_4+5H_2O_2+6H^+\rightarrow 2Z+5O_2+8H_2O$$

A
$$Mn^{2+}$$

B
$$Mn^{4+}$$

C
$$Mn$$

D
$$MnO_2$$

Solution

$$\displaystyle 2Mn{ O }^-_4+5H_2O_2+6H^+\rightarrow 2Mn^{2+}+5O_2+8H_2O $$

Hydrogen peroxide reduces $$Mn(VII)$$ to $$Mn(II)$$ in acidic medium.

Thus $$Z$$ is $$Mn^{2+}$$.

Hence, option A is correct.
Question 10
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The oxidation number of N and $$\displaystyle Cl$$ in $$\displaystyle NOCl{ O }_{ 4 }$$ respectively are :

A
+2 and +7

B
+3 and +7

C
-3 and +5

D
+2 and -7

Solution

$$\displaystyle NOCl{ O }_{ 4 }$$ is actually $$\displaystyle { NO }^{ + }{ ClO }_{ 4 }^{ - }$$.

Let the oxidation state of N in $$\displaystyle { NO }^{ + }$$ is.

$$\displaystyle x+\left( -2 \right) =+1$$

$$\displaystyle x=+1+2=+3$$

Let the oxidation state of $$\displaystyle Cl$$ in $$\displaystyle { ClO }_{ 4 }^{ - }$$ is y.

$$\displaystyle y+\left( -2 \right) \times 4=-1$$

$$\displaystyle y-8=-1;y=+7$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 12

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Redox Reactions Test - 12

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Question 1

it contains $$ Cs^{3+}$$ and $$ I^{-}$$ ions

it contains $$ Cs^{+}, I^{-}$$ and lattice $$ I_{2}$$ molecule

it is a covalent molecule

it contains $$ Cs^{+}$$ and $$ I_{3}{^-}$$ ions

Solution

Question 2

$$Cu_2I_2$$ is formed

$$CuI_2$$ is formed

$$Na_2S_2O_3$$ is oxidised

Evolved $$I_2$$ is reduced

Solution

Question 3

$$[Fe(CN)_6]^{3-}$$ and $$[Cu(CN)_4]^{2-}$$

$$MnO_2$$ and $$CrO_2Cl_2$$

$$[FeCl_4]^-$$ and $$Co_2O_3$$

$$[NiCl_4]^{2-}$$ and $$[CoCl_4]^{2-}$$

Question 4

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 5

$$\mathrm{M}\mathrm{n}\mathrm{O}_{2},\ \mathrm{F}\mathrm{e}\mathrm{C}1_{3}$$

$$[\mathrm{M}\mathrm{n}\mathrm{O}_{4}]^{-},\ \mathrm{C}\mathrm{r}\mathrm{O}_{2}\mathrm{C}1_{2}$$

$$[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})_{6}]^{3-},\ [\mathrm{C}\mathrm{o}(\mathrm{C}\mathrm{N})_{3}]$$

$$[\mathrm{N}\mathrm{i}\mathrm{C}1_{4}]^{2-},\ [\mathrm{C}\mathrm{o}\mathrm{C}1_{4}]^{-}$$

Solution

Question 6

$$HNO_3, NO, NH_4Cl, N_2 $$

$$HNO_3, NO, N_2, NH_4Cl$$

$$HNO_3, NH_4Cl, NO, N_2$$

$$NO, HNO_3, NH_4Cl, N_2$$

Solution

Question 7

heating $$\mathrm{N}\mathrm{H}_{4}\mathrm{N}\mathrm{O}_{2}$$

heating $$\mathrm{N}\mathrm{H}_{4}\mathrm{N}\mathrm{O}_{3}$$

$$Mg_{3}\mathrm{N}_{2}+\mathrm{H}_{2}\mathrm{O}$$

$$\mathrm{N}\mathrm{a}(\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}.)+\mathrm{H}_{2}\mathrm{O}_{2}$$

Solution

Question 8

$$MnO_{4}^{-} = 2, C_{2}O_{4}^{2-} = 16, H^{+} = 5$$

$$MnO_{4}^{-} = 16, C_{2}O_{4}^{2-} = 5, H^{+} = 2$$

$$MnO_{4}^{-} = 5, C_{2}O_{4}^{2-} = 16, H^{+} = 2$$

$$MnO_{4}^{-} = 2, C_{2}O_{4}^{2-} = 5, H^{+} = 16$$

Solution

Question 9

$$Mn^{2+}$$

$$Mn^{4+}$$

$$Mn$$

$$MnO_2$$

Solution

Question 10

+2 and +7

+3 and +7

-3 and +5

+2 and -7

Solution

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