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makkiz [27]
3 years ago
8

Which identifies an oxidation-reduction reaction?

Chemistry
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄  →  Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is  also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized

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2 years ago
A reaction has activation energy of 85kjper mol. What is the effect on the rate of raising the temperature from 20degree to 30 d
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Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree​

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

where k_2 = rate constant at temp T_2

k_1 = rate constant at temp T_1

E_a= activation energy

R= gas constant

T_1= temperature = 20^0C=(20+273)K=293K

T_2= temperature = 30^0C=(30+273)K=303K

ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]

ln \frac{k_{2}}{k_{1}}=1.15

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Thus rate increases 3 times on raising the temperature from 20degree to 30 degree​

3 0
2 years ago
You wish to add 5 mg/l naocl as cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that
Alecsey [184]

Answer: -

0.1 ml of bleach should be added to each liter of test solution.

Explanation:-

Let the volume of bleach to be added is B ml.

Density of stock solution = 1.0 g/ml

Mass of stock solution = Volume of stock x density of stock

                                     = B ml x 1.0 g/ml

                                     = B g

Amount of NaOCl in this stock solution = 5% of B g

                                     = \frac{5}{100} x B g

                                     = 0.05 B g

Now each test solution must be added 5 mg/l NaOCl.

Thus each liter of test solution must have 5 mg.

Thus 0.05 B g = 5 mg

                        = 0.005 g

B = \frac{0.005}{0.05}

  = 0.1

Thus 0.1 ml of bleach should be added to each liter of test solution.

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