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2 years ago

Write equations for the half-reactions that occur in the electrolysis of molten potassium bromide.

Chemistry

2 answers:

lutik1710 [3]2 years ago

4 0

The electrolysis of molten potassium bromide produces K metal at the cathode and Br gas at the anode

<h3>Further explanation</h3>

Electrolysis utilizes electrical energy to carry out non-spontaneous redox reactions. The ions in the solution flowed by an electric current will move toward the electrode opposite the charge.

The electrolyzed substance is an electrolyte which can be in the form of a melt or a solution.

In the electrolytic cell the positive pole - the anode is the site of oxidation, whereas the negative pole - the cathode is the site of the reduction.

The results of the reaction at the anode are based on substances around the anode which are easily oxidized while the reaction at the cathode is based on substances that are easily subjected to reduction.

Electrons (electricity) enter the electrolysis cell through the negative pole (cathode)

The negative ions from the solution will move towards the positive electrode, and release electrons around the positive electrode (oxidation occurs) and electrons flow to the negative pole

While around the negative electrode electron binding occurs and a reduction reaction occurs

In principle the reaction at the cathode:

1. the reduced active metal is water, other than that, the metal will be reduced

2. H⁺ of the acid will be reduced

For reactions in anode:

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

In the electrolysis of molten KBr with an inert electrode, the cation will be reduced at the cathode and the anion will be oxidized at the anode

Cathode (-): K⁺ (l) + e⁻ ⇒ K (l) (x 2)

Anode (+): 2Br⁻ (l) ⇒Br₂ (g) + 2e⁻

2K⁺ (l) + 2Br⁻9l) ⇒ 2K (l) + Br₂ (g)

metal K is produced at the cathode and Br gas at the anode

<h3>Learn more</h3>

an electrolytic cell

brainly.com/question/1319774

During electrolysis

brainly.com/question/11868966

the substance is produced at the cathode during the electrolysis

brainly.com/question/6854891

Keywords: the half-reactions, electrolysis, KBr

zzz [600]2 years ago

3 0

K⁺ + e⁻ ===> K(l)

2Br⁻ ===> Br2(g) + 2e⁻

hope that helps

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A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, <em>ΔTf </em>is the depression in freezing point of water.

<em>i</em> is van't Hoff factor.

<em>Kf </em>is the molal depression constant.

<em>m</em> is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).

van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

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2 years ago

Consider a culture medium on which only gram-positive organisms such as Staphylococcus aureus colonies can grow due to an elevat

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<h2>Selective & Differential Medium</h2>

Explanation:

Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria

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2 years ago

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Naddik [55]

So the put a lot of words to make this seem more complicated than it is. Your first equation involves the money. I’m going to use x to represent tacos and y to represent burritos.
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2 years ago

2.1. The lithium ion in a 250,00 mL sample of mineral water was

juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

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2 years ago

A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevatio

Goryan [66]

Answer:

MCl₂

Explanation:

The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.

$\Delta T=imK_b\qquad\text{where i=ions/mole, m=molality, $K_b\approx 0.512$}\\\\376.81-373.15=(x+1)\dfrac{\text{moles}}{\text{kg solvent}}(0.512)\\\\\dfrac{3.66}{0.512}=(x+1)\dfrac{\dfrac{30.2}{56+35.45x}}{0.1}=\dfrac{302(x+1)}{56+35.45x}\\\\\dfrac{3.66}{0.512\cdot 302}(56+35.45x)=x+1\\\\\dfrac{3.66\cdot 56}{0.512\cdot 302}-1=x\left(1-\dfrac{3.66\cdot 35.45}{0.512\cdot 302}\right)\\\\x=\dfrac{50.336}{24.877}\approx 2.023$

Then the formula for the salt is MCl₂.

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2 years ago