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Tems11 [23]
3 years ago
5

Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the

Chemistry
1 answer:
Leya [2.2K]3 years ago
3 0

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The balanced chemical reaction will be:

K+H_2O\rightarrow KOH+H_2

The oxidation-reduction half reaction will be :

Oxidation : K\rightarrow K^++1e^-

Reduction : 2H^++2e^-\rightarrow H_2

In this reaction, the oxidation state of 'K' changes from (0) to (+1) that means 'K' lost 1 electron and it shows oxidation and the oxidation state of 'H' changes from (+1) to (0) that means 'H' gains 1 electron and it shows reduction.

In this reaction, potassium is the oxidizing atom that has the highest oxidation state is (+1).

(b) The balanced chemical reaction will be:

Ba+2HBr\rightarrow BaBr_2+H_2

The oxidation-reduction half reaction will be :

Oxidation : Ba\rightarrow Ba^{2+}+2e^-

Reduction : 2H^++2e^-\rightarrow H_2

In this reaction, the oxidation state of 'Ba' changes from (0) to (+2) that means 'Ba' lost 2 electrons and it shows oxidation and the oxidation state of 'H' changes from (+1) to (0) that means 'H' gains 1 electron and it shows reduction.

In this reaction, barium is the oxidizing atom that has the highest oxidation state is (+2).

(c) The balanced chemical reaction will be:

Sn+I_2\rightarrow SnI_2

The oxidation-reduction half reaction will be :

Oxidation : Sn\rightarrow Sn^{2+}+2e^-

Reduction : I_2+2e^-\rightarrow 2I^-

In this reaction, the oxidation state of 'Sn' changes from (0) to (+2) that means 'Sn' lost 2 electron and it shows oxidation and the oxidation state of 'I' changes from (0) to (-1) that means 'I' gains 1 electron and it shows reduction.

In this reaction, tin is the oxidizing atom that has the highest oxidation state is (+2).

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The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

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<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

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If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

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4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

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⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

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