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MAVERICK [17]
3 years ago
7

Consider four different samples: aqueous LiBr, molten LiBr, aqueous AgBr, and molten AgBr. Current run through each sample produ

ces one of the following products at the cathode: solid lithium, solid silver, or hydrogen gas. Match each sample to its cathodic product. Drag the appropriate items to their respective bins.
Chemistry
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

  • aqueous LiBr → hydrogen gas
  • molten LiBr → solid lithium
  • Aqueous AgBr → solid silver
  • molten AgBr → solid silver

Explanation:

Electrolysis is a process in which an electric current is passed through an electrolyte in an electrolytic cell. The electrolyte maybe in <em>aqueous form or in molten state. </em>The electrolyte is an ionic compound, so during electrolysis, the anions move towards anode while the cations move towards the cathode. Oxidation takes place at the anode. While reduction takes place at the cathode.

In the molten form, only the constituent ions are present to be oxidized or reduced. Thus, in electrolysis of molten LiBr, lithium is reduced at the cathode and bromine is oxidized at the anode.

Li^{+}(aq) + e^{-} → Li(s)

Solid Lithium is produced at the cathode

Similarly, solid Silver is produced at the cathode during electrolysis of molten AgBr:

Ag^{+}(aq) + e^{-} → Ag(s)

However, in aqueous form, H_{2} O ions and LiBr ions compete with each other. Since H+ has a greater standard redox potential than lithium,H+ will be reduced at the cathode to produce hydrogen gas

2H^{+}(aq) + 2e^{-} → H_{2}(g)

Similarly,in aqueous AgBr, Ag will be produced at the cathode because the standard redox potential for Ag+ is higher than H+ .

Ag^{+}(aq) + e^{-} → Ag(s)

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For each balanced reaction, indicate the total number of atoms in the table below.
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(i) Make ammonia number of atoms on reactant side is 8 and on product side is 8. (ii) Separate water number of atoms on reactant side is 6 and on product side is 6. (iii) Combust methane number of atoms on reactant side is 9 and number of atoms on product side is 9.  

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

(i) Make Ammonia

First we have to write the balanced chemical equation

             N₂ + 3H₂ → 2NH₃

Reactant side              Product side

N = 2                             N = 2

H = 6                             H = 6

So total number of atoms on product and reactant side is 8.

(ii) Separate Water

First we have to write the balanced chemical equation

            2H₂O → 2H₂ + O₂

Reactant side              Product side

H = 4                             H = 4

O = 2                             O = 2

So total number of atoms on product and reactant side is 6.

(iii) Combust methane

First we have to write the balanced chemical equation

       CH₄ + 2O₂ → 2H₂O + CO₂

Reactant side              Product side

C = 1                             C = 1

H = 4                            H = 4

O = 4                            O = 4

So total number of atoms on product and reactant side is 9.

Thus from the above conclusion we can say that For each balanced reaction, the total number of atoms are

Reaction                            Total number of atoms

                                     Reactant side           Product side

Make Ammonia                   8                               8

Separate Water                   6                               6

Combust Methane              9                               9

Learn more about the Balanced Chemical Equation here: brainly.com/question/26694427

#SPJ1

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: For each balanced reaction, indicate the total number of atoms in the table below.

Reaction                            Total number of atoms

                                     Reactant side           Product side

Make Ammonia            _________              __________

Separate Water            _________              __________

Combust Methane       _________              __________

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Answer: 6.162g of Ag2SO4 could be formed

Explanation:

Given;

0.255 moles of AgNO3

0.155 moles of H2SO4

Balanced equation will be given as;

2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)

Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,

Therefore the number of moles of Ag2SO4 produced is given by,

n(Ag2SO4) = 0.255 mol of AgNO3 ×

[0.155mol H2SO4 ÷ 2 mol AgNO3] x

[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]

= 0.0198 mol of Ag2SO4.

mass = no of moles x molar mass

From literature, molar mass of Ag2SO4 = 311.799g/mol.

Thus,

Mass = 0.0198 x 311.799

= 6.162g

Therefore, 6.162g of Ag2SO4 could be formed

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Answer:

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On comparison

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∆ S° = 4R

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3 years ago
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