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Illusion [34]
3 years ago
8

What does the oxidizing agent do in a redox reaction apex?

Chemistry
2 answers:
densk [106]3 years ago
6 0
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)


Anni [7]3 years ago
4 0

<u>Answer:</u> It helps in the oxidation of other substance and itself gets reduced.

<u>Explanation:</u>

Oxidizing agent is defined as the agent which helps in the oxidation of other substance and itself gets reduced. It undergoes reduction reaction in any redox reaction.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Reducing agent is defined as the agent which helps in the reduction of other substance and itself gets oxidized. It undergoes oxidation reaction in any redox reaction.

Oxidation reaction is defined as the reaction in which a substance looses electrons. The oxidation state of the substance is increased.

<u>For Example:</u> Reaction of silver nitrate with copper metal, the equation follows:

Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)

The half reactions for the above reaction are:

Oxidation half reaction:  Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-

Reduction half reaction:  2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)

From the above reactions, silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

Copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Hence, it helps in the oxidation of other substance and itself gets reduced.

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irina [24]

Answer:

Concentrated solutions

7 0
2 years ago
Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha
Aleks04 [339]

Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

pH = pKa + log \frac{[Base]}{[Acid]}

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

7.5 = 6.3 + log \frac{[base]}{[acid]}

1.2 = log \frac{[base]}{[acid]}

\frac{[Base]}{[Acid]} = 10^{1.2}

\frac{[Base]}{[Acid]} = 15.8

[Base] = 15.8 \times [Acid]

The total of mole fraction of acid and base is 1. Therefore we have,

[Acid] + [Base] = 1

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

[Acid] + 15.8 \times [Acid] = 1

16.8 [Acid] = 1

[Acid] = \frac{1}{16.8}

[Acid] = 0.0595

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

4 0
3 years ago
Read 2 more answers
Li+1 had gained ______ one electron
zysi [14]
Maybe molecules one electron
3 0
3 years ago
The Lewis structure of which molecule requires resonance structures?A.) MgCl2B.) SiO2C.) SO2D.) OCl2
inna [77]

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Options (a), (b), and (d) cannot form resonating structures as they involve the formation of the chemical bonds through single sigma bond.

4 0
3 years ago
Someone pls help me I will mark you as brain
Reptile [31]
3rd one I think it is.
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2 years ago
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