Redox Reactions Test

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Redox Reactions Test - 68

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Weekly Quiz Competition

Question 1
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When a piece of sodium metal is dropped in water, hydrogen gas evolved because:

A
sodium is reduced and acts as an oxidising agent

B
water is oxidised and acts as a reducing agent

C
sodium loses electrons and is oxidised while water in reduced

D
water loses electrons and is oxidised to hydrogen

Solution

Reaction of sodium metal with water results in the evolution of hydrogen gas and formation of sodium hydroxide as:

$$2\overset { 0 }{ Na } +2\overset { +1 }{ { H }_{ 2 } } O\quad \longrightarrow 2\overset { +1 }{ Na } OH+\overset { 0 }{ { H }_{ 2 } } $$

Thus, the metal (Na) gets oxidized by lossing one electron and water gets reduced to hydrogen gas by gaining electron.
Question 2
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Arrange the following in increasing order of oxidation state of $$Ni$$.
$$K_2[Ni(CN)_4],K_2[NiF_6],Ni(CO)_4$$

A
$$Ni(CO)_4,K_2[Ni(CN)_4],K_2[NiF_6]$$

B
$$K_2[Ni(CN)_4],Ni(CO)_4,K_2[NiF_6]$$

C
$$Ni(CO)_4,K_2[NiF_6],K_2[Ni(CN)_4]$$

D
$$K_2[NiF_6],K_2[Ni(CN)_4],K_2[Ni(CN)_4]"$$

Solution

Let the oxidation number of $$Ni$$ be $$x$$.
Oxidation number of $$K=+1$$, $$CN=F=-1$$ and of $$CO=0$$.
(A) $$K_2[Ni(CN)_4]$$: $$2 \times 1+x+4 \times (−1)=0$$
$$2+x-4=0$$ or $$x=+2$$
(B) $$K_2[NiF_6]$$: $$2 \times 1+x+6 \times (−1)=0$$
$$2+x-6=0$$ or $$x=+4$$
(C) $$Ni(CO)_4]$$: $$x+4 \times 0=0$$
$$x+0=0$$ or $$x=0$$
thus the increasing order of oxidation state of $$Ni$$ is $$0<+2<+4$$ in:
$$Ni(CO)_4<K_2[Ni(CN)_4] < K_2[NiF_6]$$
Question 3
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Oxidation number of iodine in $${ IO }_{ 3 }^{ - }, { IO }_{ 4 }^{ - }, KI$$ and $$I_2$$ respectively is _____

A
-2, -5, -1 ,0.

B
+5, +7, -1 , 0.

C
+2, +5, +1 , 0.

D
-1, +1 , 0, +1.

Solution

Cosider the oxidation number of iodine as $$x$$. We know the charges on oxygen and potassium atoms are $$-2$$ and $$+1$$ respectively. The oxidation states of the compounds of iodine are calculated as follows:

$$IO_{3}^- \rightarrow x + 3(-2) = -1 \Rightarrow x = +5$$

$$IO_{4}^- \rightarrow x + 4(-2) = -1 \Rightarrow x = +7$$

$$KI \rightarrow +1 + x = 0 \Rightarrow x = -1$$

$$I_2 \rightarrow 0\Rightarrow x = 0$$
Question 4
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In the conversion of $$Br_2\ to\ Br{ O }_{ 3 }^{ - }$$, the oxidation state of $$Br$$ changes from:

A
+1 to +5

B
0 to -3

C
+2 to +5

D
0 to +5

Solution

Given: $$Br_2 \rightarrow Br{O_3}^{-}$$
Oxidation state of an element in its standard state is zero.
Therefore, for $$Br_2$$ oxidation number = 0
In $$Br{O_3}^{-}$$ let the oxidation number of $$Br$$ be $$x$$ and as for $$O$$ it is -2
Therefore $$x+3 \times (−2)=−1$$
$$x=-1+6=+5$$
$$\overset {0}{Br_2} \rightarrow \overset {+5}{Br}{O_3}^{-}$$
$$\therefore$$ In the conversion of $$Br_2\ to\ Br{O_3}^{-}$$, the oxidation state of $$Br$$ changes from 0 to +5.
Question 5
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What will be the reduction potential for the following half-cell reaction at $$298$$K?
(Given: $$[Ag^+]=0.1$$M and $$E^o_{cell}=+0.80$$V)

A
$$0.741$$V

B
$$0.80$$V

C
$$-0.80$$V

D
$$-0.741$$V

Solution

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

Nernst equation is,

$$E_{cell}=E^{0}_{cell}-\dfrac{0.0591}{n}logQ$$

Here the reaction is,

$$Ag^{+}+e^{-}\rightarrow Ag$$

$$n=1$$

$$E^{0}_{cell}=0.80V$$

Hence,

$$E_{cell}=E^{0}_{cell}-\dfrac{0.0591}{n}log\dfrac{1}{[Ag^{+}]}$$

$$E_{cell}=0.80-\dfrac{0.0591}{1}log\dfrac{1}{0.1}$$

$$=0.741V$$
Question 6
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Which of the following is a redox reaction?

A
Reaction of $$H_2SO_4$$ with $$NaOH$$

B
In atmosphere , formation of $$O_3$$ from $$O_2$$ by lightening

C
Formation of oxides of nitrogen from nitrogen and oxygen by lightening.

D
Evaporation of $$H_2O$$

Solution

A. The reaction of $$H_2SO_4$$ with $$NaOH$$
$$H_2SO_4+2NaOH \rightarrow Na_2SO_4+H_2O$$
Thus its acid-base Neutralisation reaction
B. In atmosphere , formation of $$O_3$$ from $$O_2$$ by lightening
$$3O_2 \rightarrow 2O_3$$ changes in one allotrope of O to another. Thus it's an allotropic formation
C. Formation of oxides of nitrogen from nitrogen and oxygen by lightening
$$N_2 + O_2 \rightarrow 2NO $$
here in $$N_2$$ N has oxidation number 0 and in $$NO$$ it is +2. Thus $$N_2$$ is reduced and similarly, oxidation of $$O$$ changes from 0 to -2.and thus $$O_2$$ is oxidised.
D. Evaporation of $$H_2O$$
$$H_2O(l) \rightarrow H_2O(g)$$
state changes only.
Thus, the formation of oxides of nitrogen from nitrogen and oxygen by lightning is the redox reaction where oxidation, as well as reduction, takes place, simultaneously.
Question 7
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Which of the following is the cell reaction that occurs when the following half-cells are combined?
$$I_2+2e^-\rightarrow 2I^-(1M)$$; $$E^o=+0.54$$V
$$Br_2+2e^-\rightarrow 2Br^-(1M); E^o=+1.09$$V

A
$$2Br^-+I_2\rightarrow Br_2+2I^-$$

B
$$I_2+Br_2\rightarrow 2I^-+2Br^-$$

C
$$2I^-+Br_2\rightarrow I_2+2Br^-$$

D
$$2I^-+2Br^-\rightarrow I_2Br_2$$

Solution

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.
Reduction potential of both $$I_2$$ and $$Br_2$$ is given. We have to note that more the value of reduction potential then less the element has the tendency to get reduced and hence $$I_2$$ having more reduction potential acts as anode i.e. undergoes oxidation. And $$Br_2$$ having less reduction potential than $$I_2$$ undergoes reduction and acts as cathode.

$$2I^-\rightarrow I_2+2e^-$$ (Oxidation)
$$Br_2+2e^{-}\rightarrow 2Br^-$$ (Reduction)
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$
$$2I^-+Br_2\rightarrow I-2+2Br^-$$ is net cell reaction
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$

Hence option (C) is correct.
Question 8
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$$E^o$$ values for the half cell reactions are given:
$$Cu^{2+}+e^-\rightarrow Cu^+$$; $$E^o=0.15$$V
$$Cu^{2+}+2e^-\rightarrow Cu$$; $$E^o=0.34$$V
What will be the $$E^o$$ of the half-cell: $$Cu^{+}+e^-\rightarrow Cu$$?

A
$$+0.49$$V

B
$$+0.19$$V

C
$$+0.53$$V

D
$$+0.30$$V

Solution

Given,
$$Cu^{2+}+e^{-}\rightarrow Cu^{+}; E^{0}_{1}=0.15V,\Delta G^{0}_{1},n_1=1$$

$$Cu^{2+}+2e^{-}\rightarrow Cu; E^{0}_{2}=0.34V,\Delta G^{0}_{2},n_2=2$$

$$Cu^{+}+e^{-}\rightarrow Cu; E^{0}_{3}=?V,\Delta G^{0}_{3},n_3=1$$

As $$\Delta G$$ is an extensive property,

$$\Delta G^{0}_{3}=\Delta G^{0}_{2}-\Delta G^{0}_{1}$$

$$\implies$$ $$ -n_3FE^{0}_{3}=-n_2FE^{0}_{2}+ n_1FE^{0}_{1}$$

$$-E^{0}_{3}=-2\times0.34+1\times0.15$$

$$E^{0}_{3}=0.68-0.15=0.53V$$

Hence, option C is correct.
Question 9
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In the redox reaction ,
$$Pb_3O_4 + 8HCl \rightarrow 3PbCl_2 + Cl_2 + 4H_2O$$ :

A
three numbers of $$Pb^{2+}$$ ions get oxidised to $$Pb^{4+}$$ state

B
one number $$Pb^{4+}$$ion get reduced to $$Pb^{2+}$$ and two numbers of $$Pb^{2+}$$ ions remain unchanged in their oxidation states

C
one number $$Pb^{2+}$$ ion get oxidised to $$Pb^{4+}$$ and two numbers of $$Pb^{4+}$$ ions remain unchanged in their oxidation states

D
three numbers of $$Pb^{4+}$$ ions get reduced to $$Pb^{2+}$$ state.

Solution

Reaction given:
$$Pb_3O_4+8HCl \rightarrow 3PbCl_2+Cl_2+4H_2O$$

$$Pb_3O_4$$ is a mixture of two $$PbO$$ and one $$PbO_2$$ and is written as:
$$Pb_3O_4 = 2PbO\cdot PbO_2$$

Therefore the reaction becomes:
$$2PbO.PbO_2$$ $$+ 8HCl\rightarrow3PbCl_2 + Cl_2 + 4H_2O$$

In the reaction, oxidation state of $$Pb$$ in $$PbO$$ is +2 and +4 in $$PbO_2$$.

Thus, there are two $$Pb^{2+}$$ ions in the product from PbO that remain unchanged and one $$Pb^{4+}$$ from $$PbO_2$$ that gets reduced to $$Pb^{2+}$$, resulting in total three $$Pb^{2+}$$ ions.

Hence in the reaction,

$$Pb_3O_4+8HCl \rightarrow 3PbCl_2+Cl_2+4H_2O$$

One number $$Pb^{4+}$$ion get reduced to $$Pb^{2+}$$ and two numbers of $$Pb^{2+}$$ ions remain unchanged in their oxidation states.
Question 10
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Which of the following is not an example of redox reaction ?

A
$$CuO + H_2 \rightarrow Cu + H_2O$$

B
$$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

C
$$2K + F_2 \rightarrow 2KF$$

D
$$BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$$

Solution

Redox reaction involves simultaneous oxidation and reduction of reacting species. Thus change in oxidation state will decide whether a reaction is redox or not. Thus assigning the oxidation states as:

A.$$\overset {+2}{Cu}O + \overset {0}{H_2} \rightarrow \overset {0}{Cu} + \overset {+1}{H_2}O$$: oxidation of H and reduction of Cu

B. $$\overset {+3}{Fe_2}O_3 + 3\overset {+2}{C}O \rightarrow 2\overset {0}{Fe} + 3\overset {+4}{C}O_2$$: oxidation of C and reduction of Fe

C. $$2\overset {0}{K} + \overset {0}{F_2} \rightarrow 2\overset {+1}{K}\overset {-1}{F}$$: oxidation of K and reduction of F

D. $$\overset {+2}{Ba}Cl_2 + H_2SO_4 \rightarrow \overset {+2}{Ba}SO_4 + 2HCl$$: not a redox reaction.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 68

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Redox Reactions Test - 68

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SHARING IS CARING

Question 1

sodium is reduced and acts as an oxidising agent

water is oxidised and acts as a reducing agent

sodium loses electrons and is oxidised while water in reduced

water loses electrons and is oxidised to hydrogen

Solution

Question 2

$$Ni(CO)_4,K_2[Ni(CN)_4],K_2[NiF_6]$$

$$K_2[Ni(CN)_4],Ni(CO)_4,K_2[NiF_6]$$

$$Ni(CO)_4,K_2[NiF_6],K_2[Ni(CN)_4]$$

$$K_2[NiF_6],K_2[Ni(CN)_4],K_2[Ni(CN)_4]"$$

Solution

Question 3

-2, -5, -1 ,0.

+5, +7, -1 , 0.

+2, +5, +1 , 0.

-1, +1 , 0, +1.

Solution

Question 4

+1 to +5

0 to -3

+2 to +5

0 to +5

Solution

Question 5

$$0.741$$V

$$0.80$$V

$$-0.80$$V

$$-0.741$$V

Solution

Question 6

Reaction of $$H_2SO_4$$ with $$NaOH$$

In atmosphere , formation of $$O_3$$ from $$O_2$$ by lightening

Formation of oxides of nitrogen from nitrogen and oxygen by lightening.

Evaporation of $$H_2O$$

Solution

Question 7

$$2Br^-+I_2\rightarrow Br_2+2I^-$$

$$I_2+Br_2\rightarrow 2I^-+2Br^-$$

$$2I^-+Br_2\rightarrow I_2+2Br^-$$

$$2I^-+2Br^-\rightarrow I_2Br_2$$

Solution

Question 8

$$+0.49$$V

$$+0.19$$V

$$+0.53$$V

$$+0.30$$V

Solution

Question 9

three numbers of $$Pb^{2+}$$ ions get oxidised to $$Pb^{4+}$$ state

one number $$Pb^{4+}$$ion get reduced to $$Pb^{2+}$$ and two numbers of $$Pb^{2+}$$ ions remain unchanged in their oxidation states

one number $$Pb^{2+}$$ ion get oxidised to $$Pb^{4+}$$ and two numbers of $$Pb^{4+}$$ ions remain unchanged in their oxidation states

three numbers of $$Pb^{4+}$$ ions get reduced to $$Pb^{2+}$$ state.

Solution

Question 10

$$CuO + H_2 \rightarrow Cu + H_2O$$

$$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$

$$2K + F_2 \rightarrow 2KF$$

$$BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$$

Solution

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