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Artist 52 [7]
3 years ago
10

Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one letter

ed choice that best fits each statement. A choice may be used once, more than once, or not at all in each set. The following questions refer to the reactions represented below. (A) H2SeO4(aq) + 2 Cl-(aq) + 2 H+(aq) → H2SeO3(aq) + Cl2(g) + H2O(l) (B) S8(s) + 8 O2(g) → 8 SO2(g) (C) 3 Br2(aq) + 6 OH-(aq) → 5 Br-(aq) + BrO3-(aq) + 3 H2O(l) (D) Ca2+(aq) + SO42-(aq) → CaSO4(s) (E) PtCl4(s) + 2 Cl-(aq) → PtCl62-(aq) A reaction in which the same reactant undergoes both oxidation and reduction
Chemistry
1 answer:
maksim [4K]3 years ago
6 0

Answer : The only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

Explanation :

Disproportionation reaction : It is defined as the reaction in which the same reactant undergoes both oxidation and reduction reaction. It is a redox reaction.

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 given balanced reaction is,

H_2SeO_4(aq)+2Cl^-(aq)+2H^+(aq)\rightarrow H_2SeO_3(aq)+Cl_2(g)+H_2O(l)

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which chlorine shows oxidation and selenium shows reduction.

(B) The given balanced reaction is,

S_8(s)+8O_2(g)\rightarrow 8SO_2(g)

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which sulfur shows oxidation and oxygen shows reduction.

(C) The given balanced reaction is,

3Br_2(aq)+6OH^-(aq)\rightarrow 5Br^-(aq)+BrO_3^-(aq)+3H_2O(l)

This reaction is a disproportionation reaction because in this reaction only one reactant bromine that shows both oxidation and reduction reaction.

(D) The given balanced reaction is,

Ca^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CaSO_4(s)

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of calcium and sulfate.

(E) The given balanced reaction is,

PtCl_4(s)+2Cl^-(aq)\rightarrow PtCl_6^{2-}(aq)

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of platinum and chlorine.

Hence, the only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

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Answer:

The structures are shown below.

Explanation:

When the acid reacts with water, it loses one proton (H⁺) and forms a base, which is the conjugate base of its acid.

The formal charge of an atom can be calculated by:

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Where X is the valence electrons of the neutral atom, Y is the unshared electrons, and Z is the shared electrons in the molecule.

a) When HCl deprotonates, it forms Cl⁻ as the conjugate base. The neutral atom Cl has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1 The structure is shown below in figure a.

b) When Hbr deprotonates it forms Br- as the conjugate base. The neutral atom has 7 valence electrons (X), the ion has 8 unshared electrons (Y) and none shared electrons, so FC = -1. The structure is shown below in figure b.

c) When CH3COOH loses a proton, it forms the conjugate base CH3COO⁻. The carbon as 4 valence electrons, hydrogen has 1 valence electron and oxygen has 6 valence electrons. The first carbon make simple bonds with each hydrogen and with the second carbon, and so, all the electrons are shared, and it has FC = 4 - (0 + 8/2) = 0, as so the hydrogens have FC = 1 - (0 + 2/2) = 0.

The second carbon does 1 simple bond with the first carbon, a double bond with one oxygen, and a simple bond with the other oxygen, and so doesn't have unshared electrons, and FC = 4 - (0 + 8/2) = 0.

The first oxygen does a double bond with the carbon, and so it has 4 unshared electrons, so FC = 6 - (4 + 4/2) = 0. The second oxygen does a simple bond with the carbon, and so has 5 unshared electrons, so FC = 6 - (5 + 2/2) = 0.

The structure is shown in figure c.

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