Redox Reactions Test

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Redox Reactions Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

Statement: The oxidation number of carbon in $$NaCNS$$ is $$+4$$

State whether the given statement is true or false.

A
True

B
False

C

Anomalous

D
None of these

Solution

Let $$x$$ be the oxidation number of C in $$NaCNS$$.
Since the overall

charge on the complex is $$0$$, the sum of oxidation states of all elements in

it should be equal to $$0$$.
Therefore, $$x+1+(-2) +(-3) = 0$$
or, $$x=+4$$
Hence, the oxidation number of carbon in $$NaCNS$$ is $$+4$$.

Hence, the given statement is $$\text{true}$$
Question 2
1 / -0

Corrosion of iron is :

A
redox process

B
neutrilization process

C
precipitation process

D
none of the above

Solution

Corrosion of iron is redox process.
At anode, iron is oxidized.
$$\displaystyle Fe \rightarrow Fe^+ + 2e^- \: \:\ \ \ \ \ \ \ \ \ \ \ \ E^0_{oxid} = 0.44 V $$
Reduction occurs at cathode.
$$\displaystyle 2H^+ + 1/2 O_2 +2e^- \rightarrow H_2O \: \: \ \ \ \ \ \ \ \ E^0_{red} = 1.23 V $$
Question 3
1 / -0

what is the Oxidation number of $$Fe$$ in $$\left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right]^{2+} $$ due to charge transfer from $$NO$$ to $${ Fe }^{ 2+ }$$.

A
1

B
2

C
3

D
4

Solution

Let $$x$$ be the oxidation number of iron in $${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] }^{ 2+ }$$.
Since the overall

charge on the complex is $$2+$$, the sum of oxidation states of all elements in

it should be equal to $$2+$$.
Therefore, $$x + 5(0) +1 =2$$
$$x =+1$$
Thus, oxidation number of $$Fe$$ in $$\left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] $$ is $$+1$$ due to charge transfer from $$NO$$ to $${ Fe }^{ 2+ }$$ ($$NO$$ becomes $$NO^+$$).
Question 4
1 / -0

The sum of oxidation number of all the atoms in a neutral molecule must be zero.

A
True

B
False

C
Ambiguous

D
None of these

Solution

The sum of oxidation number of all the atoms in a neutral molecule must be zero.
For example, neutral molecules such as $$O_2, P_4, O_3, S_8$$ and $$KMnO_4$$ have the sum of oxidation number of all the atoms equal to zero.
For an ion, the sum of oxidation number of all the atoms is equal to the charge on the ion.
For example, in cyanide ion $$(CN^-)$$, the sum of oxidation number of all the atoms is equal to $$-1$$.
In ammonium ion $$(NH_4)^+$$, the sum of oxidation number of all the atoms is equal to $$+1$$.
Question 5
1 / -0

Perxenate $$(XeO^{4-}_6$$) and perchromate ($$CrO_5$$) both have peroxide bonds.

A
True

B
False

C
Ambiguous

D
None of these

Solution

$$\text { Therefore, the given statement is False }$$
Question 6
1 / -0

Oxidation number of iron in brown ring complex (obtained in ring-test of $${ NO }_{ 3 }^{ - }$$) is ?
Brown ring complex: $${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] }^{ 2+ }$$

A
1

B
2

C
3

D
3.5

Solution

Let $$x$$ be the oxidation number of iron in $${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] }^{ 2+ }$$.
Since the overall

charge on the complex is $$+2$$, the sum of oxidation states of all elements in

it should be equal to $$+2$$.
Therefore, $$x+ 5(0) +1 = 2$$
or, $$x=+1$$
Thus, the oxidation number of iron in brown complex (obtained in ring-test of $${ NO }_{ 3 }^{ - }$$) $${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] }^{ 2+ }$$ is $$+1$$.
Question 7
1 / -0

The oxidation number of Ni in $$Ni(CO)_4$$ is $$+2$$.

A
True

B
False

C
Anomalous

D
None of these

Solution

Let $$x$$ be the oxidation number of $$Ni$$ in $$Ni(CO)_4$$.
Since the overall

charge on the complex is $$0$$, the sum of oxidation states of all elements in

it should be equal to $$0$$.
Therefore, $$x + 4 (0) = 0$$
Hence, $$x = 0$$.
Thus, the oxidation number of $$Ni$$ in $$Ni(CO)_4$$ is $$0$$.

Hence, the correct option is $$\text{B}$$
Question 8
1 / -0

What is the sum of oxidation number of various elements in $$HC{O}_{3}^{\ominus}$$ (bicarbonate) ion?

A
$$-1$$

B
$$0$$

C
$$+1$$

D
None of these

Solution

The sum of oxidation number of various elements in $${HCO}_3^-$$ (bicarbonate) ion is $$1+4+3(-2)=-1$$.
Question 9
1 / -0

The half cell involving the reaction $$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e \rightarrow 2Cr^{3+} (aq) + 7H_2O$$ can be represented as :

A
$$Pt, H^+ (aq), Cr_2O_7^{2-} (aq)$$

B
$$Pt, H^+ ,Cr^{3+} (aq)$$

C
$$Pt, H_2, Cr_2O_7^{2-} (aq), Cr^{3+} (aq)$$

D
$$Pt, H^+, Cr_2O_7^{2-} (aq), Cr^{3+} (aq)$$

Solution

It is reduction half cell of $$Cr_2O_7^{-2}$$ and is represented as:
$$(Pt), H^+, Cr_2O_7^{2-} (aq), Cr^{3+} (aq)$$
Question 10
1 / -0

The lowest possible oxidation state of nitrogen is $$-3$$ as in $$N^{3-}$$.

A
True

B
False

C
Ambiguous

D
None of these

Solution

The lowest possible oxidation state of nitrogen is $$-3$$ as in $$N^{3-}$$.

Nitrogen can form compounds in which oxidation state ranges from $$-3$$ to $$+5$$.

Ammonia, $$NH_3$$ and magnesium nitride, $$Mg_3N_2$$ have N in $$-3$$ oxidation state.

N has $$5$$ valence electrons. It accepts $$3$$ electrons to complete its octet.

Thus, it shows $$-3$$ oxidation state.

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Redox Reactions Test - 25

Result

Redox Reactions Test - 25

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Question 1

True

False

Anomalous

None of these

Solution

Question 2

redox process

neutrilization process

precipitation process

none of the above

Solution

Question 3

1

2

3

4

Solution

Question 4

True

False

Ambiguous

None of these

Solution

Question 5

True

False

Ambiguous

None of these

Solution

Question 6

1

2

3

3.5

Solution

Question 7

True

False

Anomalous

None of these

Solution

Question 8

$$-1$$

$$0$$

$$+1$$

None of these

Solution

Question 9

$$Pt, H^+ (aq), Cr_2O_7^{2-} (aq)$$

$$Pt, H^+ ,Cr^{3+} (aq)$$

$$Pt, H_2, Cr_2O_7^{2-} (aq), Cr^{3+} (aq)$$

$$Pt, H^+, Cr_2O_7^{2-} (aq), Cr^{3+} (aq)$$

Solution

Question 10

True

False

Ambiguous

None of these

Solution

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