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vova2212 [387]
3 years ago
12

Identify the oxidizing agent and the reducing agent for mg(s)+fe2+(aq)→mg2+(aq)+fe(s). express your answers as chemical formulas

separated by a comma.

Chemistry
2 answers:
-Dominant- [34]3 years ago
8 0

The Mg element is the reducing agent and the Fe element is the oxidizing agent

<h3>Further explanation</h3>

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 in oxidation number.

In the redox reaction, it is also known

Reducing agents are substances that experience oxidation

The oxidizing agent is a substance that is reduced

The formula for determining Oxidation Numbers in general:

1. Single element atomic oxidation number = 0. Examples of Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0

Group IA (Li, Na, K, Rb, Cs, and Fr): +1

Group IIA (Be, Mg, Ca, Sr and Ba): +2

H in compound = +1, except metal hydride compounds (Hydrogen which binds to IA or IIA groups) oxidation number H= -1, for example, LiH, MgH₂, etc.

2. Oxidation number O in compound = -2, except OF2 = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.

3 The oxidation number in an uncharged compound = 0,

Total oxidation number in ion = ion charge, Example NO₃⁻ = -1

Redox reactions are reactions that are accompanied by changes in oxidation numbers, so what must be examined is whether there are elements that experience changes in oxidation numbers in the reaction

Let's look at the reaction

Mg(s)+Fe²⁺(aq)→Mg²⁺(aq)+Fe(s)

Let see the change in the oxidation number of each element

Mg on the left = 0 (single element)

Fe²⁺ on the left = +2 ( ion charge)

Mg on the right = +2 ( ion charge)

Fe on the right = 0 (single element)

Means that the element Mg has increased oxidation number from 0 to +2 so that it experiences an oxidation reaction and acts as a reducing agent

While Fe has decreased the oxidation number from +2 to 0, so it has a reduction reaction and acts as an oxidizer

<h3>Learn more</h3>

an oxidation-reduction reaction

brainly.com/question/2973661

a reducing agent

brainly.com/question/2890416

element is reduced

brainly.com/question/4924694

Keywords: oxidation-reduction, an oxidizing agent, a reducing agent

lara31 [8.8K]3 years ago
6 0
Fe^2+ makes Mg go from 0 to 2+, Fe^2+ is the oxidizing, Mg makes Fe^2+ go from 2+ to 0, Mg is the reducing. 

Fe^2+ is a chemical symbol for ferrous in chemistry. Ferrous refers to iron with oxidation number of +2, denoted iron(II) or Fe2+.
Mg is a chemical symbol for Magnesium in chemistry.
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Answer : The mass of chlorine reacted with the phosphorus is, 53.25 grams.

Explanation :

First we have to calculate the moles of phosphorus.

\text{Moles of phosphorus}=\frac{\text{Mass of phosphorus}}{\text{Molar mass of phosphorus}}

\text{Moles of phosphorus}=\frac{15.5g}{31g/mol}=0.5mol

Now we have to calculate the moles of Cl_2

The balanced chemical reaction is:

2P+3Cl_2\rightarrow 2PCl_3

From the balanced chemical reaction, we conclude that

As, 2 moles of phosphorous react with 3 moles of Cl_2

So, 0.5 moles of phosphorous react with \frac{3}{2}\times 0.5=0.75 moles of Cl_2

Now we have to calculate the mass of Cl_2

\text{Mass of }Cl_2=\text{Moles of }Cl_2\times \text{Molar mass of }Cl_2

Molar mass of Cl_2 = 71 g/mol

\text{Mass of }Cl_2=0.75mol\times 71g/mol=53.25g

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If a buffer is composed of 34. 63 ml of 0. 139 m acetic acid and 36. 50 ml of 0. 182 m sodium acetate, how many ml of 0. 100 m h
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The volume is the product of the molar concentration and the moles of the substance. The volume of hydrochloric acid that can be added to the buffer is 0.066 L.

<h3>What is a buffer?</h3>

A buffer is a constant solution with a proton or hydrogen ion concentration with an acid, base, or salt dissolved in them.

The chemical reaction between hydrochloric acid and sodium acetate is given as,

HCl + CH₃Coo-Na → CH₃CooH + NaCl

From above the initials moles of acetic acid are given as,

Moles = molarity × volume

= 0.139 M × 0.0346 L

=  0.0048 moles

The initials moles of sodium acetate are given as,

M = 0.182 M × 0.0365 L

= 0.0066 moles

As it is known that the ratio of the buffer capacity is 10 for sodium acetate and acetic acid so,

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The volume of 0.100 HCl with 0.66 moles is:

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= 0.066 L

Therefore, 0.066 L of 0.100 M HCl is required to reach the buffer capacity.

Learn more about buffer here:

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