Redox Reactions Test

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Redox Reactions Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

The oxidation of toluene with $$CrO_3$$ in the presence of $$(CH_3CO)_2O$$ gives a product A, which on treatment with aqueous $$NaOH$$, produces?

A
$$C_6H_5COONa$$

B
$$2, 4-diacetyl \, toluene$$

C
$$C_6H_5CHO$$

D
$$(C_6H_5CO)_2O$$

Solution
Question 2
1 / -0

A gas X at 1 atm is bubbled through the solution containing a mixture of 1MY and 1MZ at 25 degrees. If the order of reduction potential is Z > Y > X, then

A
Y will oxidise X not Z

B
Y will oxidise Z not X

C
Y will oxidise both X and Y

D
Y will reduce both X and Y

Solution

The reduction tendencies is $$Z > Y > X$$.Hence,Y can oxidise X and not Z
Question 3
1 / -0

The complex compound in which metal is not present in zero oxidation state?

A
$${ Mn }_{ 2 }{ \left( CO \right) }_{ 10 }$$

B
$$\left[ Ni{ \left( CO \right) }_{ 4 } \right]$$

C
$$\left[ Cr{ \left( { C }_{ 6 }{ H }_{ 6 } \right) }_{ 2 } \right]$$

D
$$K\left[ { PtCl }_{ 3 }\left( { C }_{ 2 }{ H }_{ 4 } \right) \right]$$

Solution

$$A.$$ $$Mn_2{(CO)}_{10}\Longrightarrow 2x+(10\times 0)=0\Longrightarrow x=0,$$ ($$x=$$oxidation state of metal.)
$$B.$$ $$[Ni{(CO)}_{4}]\Longrightarrow x+(4\times 0)=0\Longrightarrow x=0,$$ ($$x=$$oxidation state of metal.)
$$C.$$ $$[Cr{(C_6H_6)}_{2}]\Longrightarrow x+(2\times 0)=0\Longrightarrow x=0,$$ ($$x=$$oxidation state of metal.)
$$D.$$ $$K[PtCl_3{(C_2H_4)}]\Longrightarrow +1+x+3\times (-1)+0=0\Longrightarrow x=+2,$$ ($$x=$$oxidation state of metal.)
$$\therefore$$ $$Pt$$ is in $$+2$$ Oxidation state.
$$\therefore$$ Option $$D.$$ is the correct answer.
Question 4
1 / -0

From the following half-cell reaction and their potential, What is the smallest possible standard e.m.f spontaneous reaction?
$${ PO }_{ 4 }^{ 3- } (aq) + 2H_2O(I) + 2e^- \longrightarrow { HPO }_{3}^{2-} + 3OH^-(aq); E^o = -1.05 V$$
$$ PbO _2(s) + H_2O(I) + 2e^- \longrightarrow PbO(S) + 2OH^-(aq); E^o = +0.28 V$$
$$IO{-}{3}(aq) + 2H_2O(I) + 4e^- \longrightarrow IO^-(aq) + 4OH^-(aq); E^o = +0.56 V$$

A
$$+0.00$$

B
$$+0.74$$

C
$$+0.56$$

D
$$+0.28$$

Solution
Question 5
1 / -0

In which of the following compounds, the metal is in the lower oxidation state

A
$$ [Co( NH_{3})_{5}Br_{2}]SO_{4}$$

B
$$Fe_{3} [Fe( CN)_{6}]_{2}$$

C
$$[Mn_{2}(CO)_{10}]$$

D
$$K[ PtCI_{3}(C_{2} H_{4})]$$
Question 6
1 / -0

$$H_2SO_4$$ acts as a strong oxidising agent. In which of the reaction, it is not acting as an oxidising agent?

A
$$C + 2H_2SO_4 \rightarrow CO_2 + 2SO_2 + 2H_2O$$

B
$$CaF_2 + 2H_2SO_4\rightarrow CaSO_4 + 2HF$$

C
$$S + 2H_2SO_4 \rightarrow 3SO_2 + H_2O$$

D
$$Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O$$
Question 7
1 / -0

Which is not a redox reaction?

A
$$BaO_{2}+H_{2}SO_{4}\rightarrow BaSO_{4}+H_{2}O_{2}$$

B
$$2BaO_+O{_2}\rightarrow 2BaO_{2}$$

C
$$4KClO_{3}\rightarrow 4KClO_{2}+2O_{2}$$

D
$$SO_{2}+2H_{2}S\rightarrow 2H_{2}O+3S$$

Solution

Here $$BaO_2+H_2SO_4\longrightarrow BaSO_4+H_2O_2$$ is a double displacement reaction where the cations and anions of the $$2$$ reactants switch places forming $$2$$ new compounds.
All other options involve change in oxidation status. Hence they are redox reactions.
Question 8
1 / -0

Fe shows an oxidation state of $$+1$$ in:

A
$$Fe(CO)_5$$

B
$$[Fe(H_2O)_5 NO]SO_4$$

C
$$Fe_4[Fe(CN)_6]$$

D
$$FeCl_4^-$$

Solution

In the standard brown ring test for the nitrate ion, the brown ring complex is:
$$[Fe(H_2O)_5(NO)]^{2+}$$
In this compound, the nitrosyl ligand is positively charged, and iron is in a +1 oxidation state.
Hence option B is correct answer.
Question 9
1 / -0

What is the oxidation number of $$Cl$$ in the $$ClO^{-}_3$$ ion?

A
-1

B
+5

C
-5

D
+7

Solution

The oxidation number of chlorine can be $$-1,0,+1,+3,+4,+5,or+7$$ depending on the substance containing the chlorine.
So,
In $$Cl{{O}_{{{3}^{-}}}}$$ if $$O=- 2$$

Then oxidation state of $$Cl=+5$$
Question 10
1 / -0

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

A
$$ CH_4 $$

B
$$ NH_3 $$

C
$$ HCI $$

D
$$ CaH_2 $$

Solution

In metallic hydrides the oxidation number of hydrogen is -1.

A metal hydride has hydrogen bonded to a metal.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 54

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Redox Reactions Test - 54

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SHARING IS CARING

Question 1

$$C_6H_5COONa$$

$$2, 4-diacetyl \, toluene$$

$$C_6H_5CHO$$

$$(C_6H_5CO)_2O$$

Solution

Question 2

Y will oxidise X not Z

Y will oxidise Z not X

Y will oxidise both X and Y

Y will reduce both X and Y

Solution

Question 3

$${ Mn }_{ 2 }{ \left( CO \right) }_{ 10 }$$

$$\left[ Ni{ \left( CO \right) }_{ 4 } \right]$$

$$\left[ Cr{ \left( { C }_{ 6 }{ H }_{ 6 } \right) }_{ 2 } \right]$$

$$K\left[ { PtCl }_{ 3 }\left( { C }_{ 2 }{ H }_{ 4 } \right) \right]$$

Solution

Question 4

$$+0.00$$

$$+0.74$$

$$+0.56$$

$$+0.28$$

Solution

Question 5

$$ [Co( NH_{3})_{5}Br_{2}]SO_{4}$$

$$Fe_{3} [Fe( CN)_{6}]_{2}$$

$$[Mn_{2}(CO)_{10}]$$

$$K[ PtCI_{3}(C_{2} H_{4})]$$

Question 6

$$C + 2H_2SO_4 \rightarrow CO_2 + 2SO_2 + 2H_2O$$

$$CaF_2 + 2H_2SO_4\rightarrow CaSO_4 + 2HF$$

$$S + 2H_2SO_4 \rightarrow 3SO_2 + H_2O$$

$$Cu + 2H_2SO_4 \rightarrow CuSO_4 + SO_2 + 2H_2O$$

Question 7

$$BaO_{2}+H_{2}SO_{4}\rightarrow BaSO_{4}+H_{2}O_{2}$$

$$2BaO_+O{_2}\rightarrow 2BaO_{2}$$

$$4KClO_{3}\rightarrow 4KClO_{2}+2O_{2}$$

$$SO_{2}+2H_{2}S\rightarrow 2H_{2}O+3S$$

Solution

Question 8

$$Fe(CO)_5$$

$$[Fe(H_2O)_5 NO]SO_4$$

$$Fe_4[Fe(CN)_6]$$

$$FeCl_4^-$$

Solution

Question 9

-1

+5

-5

+7

Solution

Question 10

$$ CH_4 $$

$$ NH_3 $$

$$ HCI $$

$$ CaH_2 $$

Solution

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