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2 years ago

PLZ HELP, Its The Nature of Oxidation and Reduction for chemistry

Chemistry

1 answer:

elixir [45]2 years ago

8 0

Answer:

Oxidation is defined as the chemical process in which substance loses electron and hydrogen or gain oxygen while in the process of reduction, substance gains electron and hydrogen or loses oxygen.

So, from the given equation:

a. It is an oxidation reaction as Rb loses one elctron.

b. It is a reduction reaction as Te gains two electrons and become Te2-

c. It is a reduction reaction as H atom gains electrons.

d. It is an oxidation reaction as P loses 3 electrons.

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Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago

The following equilibrium is formed when copper and bromide ions are placed in a solution:

JulsSmile [24]

Answer:

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

$heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)$

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0

3 years ago

Why doesn't gravity pull water all the way to the center of the Earth?

ElenaW [278]

It has to do with force

6 0

3 years ago

Please help!!i NEED to get this right

Lana71 [14]

The correct answer is C

3 0

2 years ago

Read 2 more answers

Two molecules of hydrogen react with one molecule of oxygen to produce two

Alex787 [66]

You would need to utilize Mole ratios found in the adjusted condition;
for each mole of hydrogen utilized, 2 moles of HCl are delivered.
Thusly:
10 mol H2 x 2 mol HCl/1 mol H2 = 20 mol HCL.
For the second question:
you would need to change over 2.0x10^23 particles of Oxygen to moles of oxygen, utilizing Avogadro's number:
2.0x10^23 particles oxygen x 1 mol oxygen/6.022x10^23 atoms oxygen = 0.33 mol Oxygen
utilizing mole proportions once more:
0.66 mol H2O = 2 mol H2O/1 mol Oxygen x 0.33 mol Oxygen
45.0 mol H2O = 2 mol H2O/1 mol Oxygen x 22.5 mol Oxygen
fundamentally to answer stoichiometry, you should take a gander at the adjusted condition to make sense of the mole proportions between components/mixes, and utilizing mole proportions you can change over from moles of one component/compound to moles of another component/compound

4 0

3 years ago

Read 2 more answers