Redox Reactions Test

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Redox Reactions Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Oxidation number of $$Cr$$ in the following complex is:

A
3

B
6

C
4

D
5

Solution

Oxidation state of $$Cr$$ in $$Cr(H_2O)_6^{3+}$$ is the charge on the ion= $$+3$$
As $$H_2O$$ units do not contribute to the charge on the ion, the only atom that contribute is the $$Cr$$ atom.
Question 2
1 / -0

For the half reaction
$$B(s) \rightarrow B^{2+} + 2e^{-}$$ $$E_{1}^{\circ} = - 0.44 V$$
and $$B_{2}+\rightarrow B^{3+} + e^{-}$$ $$E_{2}^{\circ} = 1.3 V$$
What is $$E^{\circ}$$ for the reaction,
$$3e^{-} + B^{+3} \rightarrow B(s)$$

A
$$0.86$$ V

B
$$0.14$$ V

C
$$-0.14$$ V

D
$$-0.28$$ V

Solution

$$B(s)\rightarrow { B }^{ 2 }+{ 2e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=-0.44V$$ $$- i$$
$${ B }^{ 2+ }\rightarrow { B }^{ 3+ }+{ e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=1.3V$$ $$- ii$$
$${ B }^{ 3+ }+{ 3e }^{ - }\rightarrow B(s)\quad \quad \quad \quad { E }_{ 3 }^{ 0 }=?$$ $$- iii$$
iii $$=- (II+I)$$
Or, $$-3\times { E }_{ 3 }^{ 0 }\times F=-\left\{ -1\times 1.3\times F+0.44\times 2\times F \right\} $$
$$\therefore { E }_{ 3 }^{ 0 }=-0.14V$$
Question 3
1 / -0

The oxidation number of S is $$(CH_{3})_{2}SO$$ is ?

A
1

B
2

C
0

D
3

Solution

1)Structure: $$S$$ is double-bonded to$$ O$$ ($$— 2$$ electrons) and is bonded to $$2 C$$ ($$+2$$ electrons).
2)$$ +2 + -2 = 0. S$$has a$$ 0 $$Oxidation number in $$(CH_3)_2SO$$.
Question 4
1 / -0

The oxidation number of S in $$(CH_{3})_{2}SO$$ is ?

A
1

B
2

C
0

D
3

Solution

$$S$$ is double-bonded to$$ O (— 2$$ electrons$$) $$and is bonded to $$2 C (+2$$ electrons).
$$+2 + -2 = 0$$.
$$S$$ has a$$ 0 $$Oxidation number in $$(CH_3)_2SO$$.
Question 5
1 / -0

The standard electrode potential of the half cells is given below:
$$Zn^{2+}+2e^{-}\rightarrow Zn; E=-0.76 V$$
$$Fe^{2+}+2e^{-}\rightarrow Fe; E=-0.44 V$$
The emf of the cell $$Fe^{2+}=Zn^{-}\rightarrow Zn^{2+}+Fe $$

A
$$1.54 V$$

B
$$-1.54 V$$

C
$$0.32V$$

D
$$+0.19 V$$

Solution

$${ E }_{ cell }^{ 0 }=\left( { E }_{ { Fe }^{ 2+ }|Fe }^{ 0 }+{ E }_{ Zn|{ Zn }^{ 2+ } }^{ 0 } \right) $$
$$=\left( -0.44 V + 0.76 \right)$$
$$E^0=+0.32 Volt$$
Question 6
1 / -0

Highest positive oxidation state shown by:

A
Cl

B
Os

C
Mn

D
Cr

Solution

Osmium exhibits oxidation states from $$0 - +8 $$ in its compounds, with the exception of $$+1$$.
well-characterized and stable compounds contain the element in$$ +2, +3, +4, +6,$$ and $$+8$$ states.
There are also carbonyl and organometallic compounds in the low oxidation states $$−2, 0, +1$$.
Question 7
1 / -0

The lowest oxidation state shown by an element in its compound is ?

A
$$-5$$

B
$$-3$$

C
$$-4$$

D
$$-2$$

Solution

The state of an element or ion in a compound with regard to the electrons gained or lost by the element or ion in the reaction that formed the compound, expressed as a positive or negative number indicating the charge of the ion is defined as the oxidation state.
The lowest oxidation state is $$-4,$$ as for carbon in methane or for chromium in $$[Cr(CO)_4]^{4-}$$
Question 8
1 / -0

The elements having lowest and highest oxidation number respectively are:

A
$$Sc,Fe$$

B
$$Sc,Mn$$

C
$$Ag,Mn$$

D
$$Fe,Mn$$

Solution
Question 9
1 / -0

The conversion of 2,3-dibromobutane to 2-butene with $$Zn$$ is:

A
redox reaction

B
$$\alpha - $$elimination

C
dehalogenation

D
both $$\alpha $$ elimination and redox reaction

Solution

Heating of $$2,3-dibromobutane$$ with $$Zn$$ dust yields $$2-butene$$ as shown below.
$$CH_3-{\underset {|}CH}-\underset {|}{CH}-CH_3\xrightarrow []{\text {Zn dust}}CH_3-CH=CH-CH_3$$
$$Br$$ $$Br$$
It is a dehalogenation reaction.
Question 10
1 / -0

Oxidation numbers of P in $$P{ O }_{ 4 }^{ 3- }$$, of S in $$S{ O }_{ 4 }^{ 2- }$$ and that of Cr in $$C{ r }_{ 2 }{ O }_{ 7 }^{ 2- }$$ are respectively:

A
+3,+6 and +5

B
+5,+3and +6

C
-3,+6 and +6

D
+5,+6 and +6

Solution

(I) $$^xPO_4^{3-} \Rightarrow x+4\times (-2)=-3$$
$$\Rightarrow x= -3+8=+5$$
$$\Rightarrow x=+5$$
Oxidation number of $$P=+5$$
(II) $$^xSO_4^{2-} \Rightarrow x+4\times (-2)= -2$$
$$\Rightarrow x= -2+8$$
$$\Rightarrow x=+6$$
Oxidation number of $$S=+6$$
(III) $$^x{Cr_2}O_7^{2-} \Rightarrow 2x+7\times (-2)=-2$$
$$\Rightarrow 2x=-2+14$$
$$\Rightarrow 2x=12$$
$$\Rightarrow x=\cfrac {12}{2}=+6$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 53

Result

Redox Reactions Test - 53

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SHARING IS CARING

Question 1

3

6

4

5

Solution

Question 2

$$0.86$$ V

$$0.14$$ V

$$-0.14$$ V

$$-0.28$$ V

Solution

Question 3

1

2

0

3

Solution

Question 4

1

2

0

3

Solution

Question 5

$$1.54 V$$

$$-1.54 V$$

$$0.32V$$

$$+0.19 V$$

Solution

Question 6

Cl

Os

Mn

Cr

Solution

Question 7

$$-5$$

$$-3$$

$$-4$$

$$-2$$

Solution

Question 8

$$Sc,Fe$$

$$Sc,Mn$$

$$Ag,Mn$$

$$Fe,Mn$$

Solution

Question 9

redox reaction

$$\alpha - $$elimination

dehalogenation

both $$\alpha $$ elimination and redox reaction

Solution

Question 10

+3,+6 and +5

+5,+3and +6

-3,+6 and +6

+5,+6 and +6

Solution

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