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3 years ago

What happens to a reducing agent during a redox reaction?

Chemistry

2 answers:

d1i1m1o1n [39]3 years ago

8 0

The reducing agent is oxidized during a redox reaction

$\boxed{\boxed{\bold{Further~explanation}}}$

The oxidation-reduction reaction or abbreviated as Redox is a chemical reaction in which there is a change in oxidation number

3 basic theories explain this Redox concept:

1. Binding/release of oxygen

The oxidation reaction is the binding of a substance with oxygen. (O₂)

For example:

2SO₂ + O₂ ----> 2SO₃

The reduction reaction is the release of oxygen from a substance.

For example:

2CuO → 2Cu + O₂

2. Electron release / binding reaction

Oxidation is an electron release event

Example:

2F ---> 2Fe³⁺ + 6e⁻

The reduction is an electron capture event

Example:

3O₂ + 6e⁻ ---> 3O²⁻

3. The reaction of addition/reduction of oxidation number

Oxidation is an increase/increase in oxidation number, while reduction is a decrease/decrease in oxidation number.

In this reaction, it is also known

Reducing agents are substances that experience oxidation

The oxidizer is a substance that is reduced

Let's look at the answer choices

The reducing agent is reduced.

This statement is wrong because the reducing agent undergoes oxidation

The reducing agent is oxidized.

This statement is true

The reducing agent remains unchanged.

The statement is wrong because the oxidation number of the oxidized compound will increase

The reducing agent gains electrons.

The statement is wrong because the reducing agent releases electrons

So the correct statement is B The reducing agent is oxidized

$\boxed{\boxed{\bold{Learn~more}}}$

chemical reactions are an oxidation-reduction reaction

brainly.com/question/2973661

the reduction process

brainly.com/question/1222373

an element's oxidation

brainly.com/question/3775501

an element's oxidation number increase

brainly.com/question/6068719

Keywords: oxidation, reduction, The oxidation number

Anarel [89]3 years ago

3 0

During a redox reaction, a reducing agent loses electron and become oxidised. The correct option is B.

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A solution of h2so4(aq) with a molal concentration of 2.24 m has a density of 1.135 g/ml. what is the molar concentration of thi

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This problem is simply converting the concentration from molality to molarity. Molality has units of mol solute/kg solvent, while molarity has units of mol solute/L solution.

2.24 mol H2SO4/kg H2O * (0.25806 kg H2SO4/mol H2SO4) = 0.578 kg H2SO4/kg H2O

That means the solution weighs a total of 1 kg + 0.578 kg = 1.578 kg. Then, convert it to liters using the density data:

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3 years ago

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Explanation:

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3 years ago

Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the d

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Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

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3 years ago

1. How many molecules of S2 gas are in 756.2 L?

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Answer: There are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

Explanation:

It is known that 1 mole of any gas equals 22.4 L at STP. Hence, number of moles present in 756.2 L are calculated as follows.

$Mole = \frac{Volume}{22.4 L}\\= \frac{756.2 L}{22.4 L}\\= 33.76 mol$

According to mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ molecules.

Therefore, molecules of S present in 33.76 moles are calculated as follows.

$1 mol = 6.022 \times 10^{23}\\33.76 mol = 33.76 \times 6.022 \times 10^{23}\\= 2.032 \times 10^{25}$

Thus, we can conclude that there are $2.032 \times 10^{25}$ molecules $S_{2}$ gas are in 756.2 L.

5 0

3 years ago