Electron Transfer Reactions (DP IB Chemistry)

Half-equations separate the processes of oxidation and reduction, showing the loss or gain of electrons.

The balanced half equation acidic conditions is:

Cl₂ → 2Cl^- + 2e^-

The balanced half equation in acidic conditions is:

H₂O₂ → O₂ + 2H⁺ + 2e^-

The balanced half equation in acidic conditions is:

H₂S → S + 2H⁺ + 2e^-

The balanced half equation in acidic conditions is:

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

False.

The following half equation is not correctly balanced.

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

The balanced half equation in basic conditions is:

2OH¯ + 2NH₃ → N₂H₄ + 2H₂O + 2e¯

The balanced half equation in basic conditions is:

8e¯ + 4H₂O + BrO₄¯ → Br¯ + 8OH¯

The balanced half equation in basic conditions is:

2e¯ + 2H₂O + PbO₂ → PbO + H₂O + 2OH¯

True.

The following half equation is correctly balanced.

Fe + 3OH¯ → Fe(OH)₃ + 3e¯

The relative ease of oxidation increases as you move down Group 1.

False.

The reaction between potassium and water is more vigorous than the reaction between lithium and water.

The reactivity with water increases as you move down Group 2.

The oxidation number of lithium changes from 0 to +1 when it reacts with water.

The oxidising power of halogens decreases down Group 17.

The symbol equation for the reaction between chlorine and potassium bromide.

2KBr (aq) + Cl₂ (aq) → 2KCl (aq) + Br₂ (aq)

False.

Fluorine is the strongest oxidising agent among the halogens.

True.

In the following reaction Mg has displaced Cu in CuSO₄.

Mg (s) + CuSO₄ (aq) → MgSO₄  (aq) +  Cu (s)

True.

In a metal displacement reaction, the more reactive metal is always oxidised.

The half-equation for the reduction of copper(II) ions is:

Cu²⁺ + 2e^- → Cu

The general equation for the reaction between an acid and a metal is:

acid + metal → salt + hydrogen

A salt is an ionic compound formed when the hydrogen of an acid is replaced by a metal or another positive ion.

False.

Not all metals react with acids. Very unreactive metals like gold do not react with acids to produce hydrogen gas.

The symbol equation for the reaction between hydrochloric acid and zinc is:

2HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g)

The ionic equation for the reaction between zinc and hydrochloric acid is:

2H⁺ (aq) + Zn (s) → Zn²⁺ (aq) + H₂ (g)

True.

The strength of an acid affects the rate of its reaction with a metal. Stronger acids react more vigorously.

The symbol equation for the reaction between aluminium and sulfuric acid is:

2Al (s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ + 3H₂

The half equation for the oxidation of Al to Al³⁺ is:

Al → Al³⁺ + 3e^-

The change in oxidation number of hydrogen in the following equation is +1 to 0.

2Al (s) + 6HCl (aq) → 2AlCl₃ + 3H₂

The position of a metal in the reactivity series is determined by its ability to lose electrons and form positive ions.

A primary (voltaic) electrochemical cell converts energy from spontaneous redox reactions to electrical energy..

Electromotive force (EMF) is the potential difference generated by an electrochemical cell, also known as cell potential.

Electrons flow from the anode to the cathode in a primary cell.

X is the salt bridge.

X is the voltmeter.

Oxidation takes place at the Zn electrode as it is the more reactive metal / has a more negative  E^ꝋ value.

Suitable electrolytes for both electrodes could be zinc nitrate and silver nitrate (any soluble salt).

True.

A salt bridge is often a piece of filter paper saturated with a solution of an inert electrolyte such as KNO₃ (aq).

The conventional representation for the zinc-copper voltaic cell is:

Zn (s) ∣ Zn²⁺ (aq) ∥ Cu²⁺ (aq) ∣ Cu (s)

A salt bridge indicated by ∥ in a cell representation.

Secondary cells involve reversible redox reactions that can be reversed using electrical energy.

The negative electrode is lead metal and the positive electrode is lead(IV) oxide.

The general advantages of secondary cells are:

• Materials can be regenerated

• Can deliver high current

The general advantages of primary cells include they:

• Are inexpensive

• Are lightweight

• Have a long shelf life

Two advantages of fuel cells could be:

• Reduced pollution if hydrogen is used as the fuel

• Hydrogen has a low density

• More efficient than combustion as more chemical energy converted to electrical energy

Two disadvantages of fuel cells could be:

• Safety issues with hydrogen gas

• Hydrogen must be transported using heavy containers

• Expensive

• Only delivers small currents

• Technical issues due to catalytic failures, leaks and corrosion

The disadvantages of lead-acid batteries are:

• Heavy mass

• Lead and sulfuric acid could cause pollution

An advantage of using lead-acid batteries is that they can deliver large amounts of energy over short periods.

The reaction for the discharge of the nickel-cadmium cell is:

2NiO(OH) (s) + 2H₂O (l) + Cd (s) → 2Ni(OH)₂ (s) + Cd(OH)₂ (s)

False.

Cobalt in CoCO₂ is reduced during the discharge of a lithium ion cell.

 Li (s)  + CoO₂ (s)  →   Li ⁺ (CoO₂) ^– (s)   

An electrolytic cell is an electrochemical cell that converts electrical energy to chemical energy by bringing about non-spontaneous reactions.

True.

Molten ionic compounds can conduct electricity.

During electrolysis, oxidation (loss of electrons) occurs at the anode.

During electrolysis, reduction (gain of electrons) takes place at the cathode

Electrons flow from the anode to the cathode.

Positive cations are attracted to the negatively charged cathode because opposite charges attract.

Negative anions are attracted to the positively charged anode because opposite charges attract.

False.

The electrolyte must be molten or in solution to undergo electrolysis.

False.

A DC power supply is used for electrolytic cells.

The products of the electrolysis of molten zinc chloride would be zinc at the cathode and chlorine at the anode.

False.

The cathode product will always be the metal and the product formed at the anode will always be the non-metal.

During the electrolysis of lead(II) chloride, two chloride ions are attracted to the anode where they each lose one electron.

They bond to form a molecule of chlorine.

2Cl^- ⟶ Cl₂ + 2e^-

The products of the following reaction is:

CH₃CH(OH)CH₂CH₃ + [O] → CH₃COCH₂CH₃ + H₂O

False.

Secondary alcohols can only be oxidised to ketones.

The following alcohol can not be oxidised because it is a tertiary alcohol.

The two possible organic products from the reaction between propan-1-ol and acidified potassium dichromate(VI) are propanal and propanoic acid.

Propan-2-ol is a secondary alcohol.

The organic product formed when propan-1-ol undergoes reflux with acidified potassium dichromate(VI) is:

Distillation apparatus is used to form propanal from propan-1-ol using acidified potassium dichromate(VI)

The two equations to show the oxidation of butan-1-ol to butanoic acid is:

CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O

CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH

The organic product formed during the oxidation of CH₃CH(OH)CH₂CH₂CH₃ is pentan-2-one.

False.

Distillation apparatus is used to form an aldehyde from a primary alcohol.

The equation to show the reduction of propanoic acid to propan-1-ol is:

CH₃CH₂COOH +4[H] → CH₃CH₂CH₂OH + H₂O

True.

Ketones are reduced to secondary alcohols.

Carboxylic acids are reduced to aldehydes and primary alcohols.

Ketones are reduced to secondary alcohols.

The equation for the reduction of propanone to propan-2-ol is:

CH₃COCH₃ +2[H] → CH₃CH(OH)CH₃

The equation for the reduction of butanone is:

CH₃CH₂COCH₃ +2[H] → CH₃CH₂CH(OH)CH₂ + H₂O

The compound that is formed from the reduction of pentan-2-one is pentan-2-ol.

True.

Reduction of unsaturated compounds by the addition of hydrogen lowers the degree of unsaturation.

The completed equation is:

CHCH + H₂ → CH₂CH₂

The completed equation is:

CHCH + 2H₂ → CH₃CH₃

The completed equation is:

CH₂CH₂ + H₂ → CH₃CH₃

True.

Alkynes can be reduced to alkenes and alkanes.

The reaction between an alkene and hydrogen is known as hydrogenation or reduction.

False.

The following reaction is not correct.

CHCCH₃ + 2H₂ → CH₃CH₂CH₃

The standard hydrogen electrode is a half-cell used as a reference electrode to measure electrode potentials.

Hydrogen gas at a pressure of 100 kPa is used in the standard hydrogen electrode.

X is made from ‎ ‎platinum‎.

HCl (aq) with a concentration of 1.00 mol dm^-3 is used as the electrolyte in the standard hydrogen electrode.

True.

By convention the standard electrode potential of the standard hydrogen electrode is 0.00 V.

Standard electrode potential is represented by E^⦵.

True.

The more negative the half-cell, the better it can act as a reducing agent.

False.

If a hydrogen electrode is used to measure the electrode potentials of zinc, the conventional cell diagram would be:

Pt 丨H₂ (g), 2H⁺ (aq)  ∥  Zn²⁺ (aq), Zn (s)

The more ‎positive the half-cell, the better it can act as a oxidising agent.

The equation to calculate E^θ_cell is:

E^θ_cell= E_right - E_left

The conditions used when measuring standard electrode potential are:

• 1.0 mol dm^-3 ions concentrations

• 100 kPa pressure

• 298 K

To calculate the E^θ_cell for Mg (s) | Mg²⁺ (aq) || Ag⁺ (aq) | Ag (s):

E^θ_cell = +0.80 - (-2.37) = 3.17 V

To calculate the E^θ_cell for Fe (s) ⏐ Fe²⁺ (aq) ⏐⏐ Cu²⁺ (aq) ⏐ Cu (s):

E^θ_cell = +0.34 - (-0.41) = +0.75 V

True.

More positive E^ꝋ values indicate that the species is easily reduced.

False.

E^θ_cell = E_reduction – E_oxidation

The minimum value the standard electrode potential, E^θ_cell, should be for a reaction to be spontaneous is 0.00 (V).

True.

The following reaction is spontaneous because E^θ _cell  is a positive value.

Ni (s) + Cu²⁺ (aq) → Ni²⁺ (aq) + Cu (s)   E^θ _cell = +0.60 V

The following reaction is not spontaneous.

Pb (s) + Mg²⁺ (aq)  → Pb²⁺ (aq) + Mg (s)

E^θ _cell = -2.37 - (-0.13) = -2.24

The better reducing agent is Mg²⁺ (aq) as it is less positive / more negative.

n = number of electrons transferred in ΔG^θ = -nFE^θ

F = the Faraday constant (96 500 C mol^-1 ) in ΔG^θ = -nFE^θ .

True.

If ΔG^θ and E^θ are both positive the reaction is spontaneous.

False.

If ΔG^θ and E^θ are both 0 the reaction is at equilibrium.

If the reaction is non-spontaneous, ΔG^θ is positive and E^θ is negative.

To calculate ΔG^θ :

• ΔG^θ = -nFE^θ

• ΔG^θ = -2 x 96 500 x 0.60 V = -115800 (J mol^-1)

For the following reaction the value for n in ΔG^θ = -nFE^θ is 2.

Zn (s) +  Cu²⁺ (aq) → Zn²⁺ (aq)  +  Cu (s)

For the following reaction the value for n in ΔG^θ = -nFE^θ is 1.

Cu⁺ (aq) → Cu (s) + Cu²⁺ (aq)

During the electrolysis of water the product at the anode is oxygen.

During the electrolysis of water, the product at the anode is hydrogen.

The gas that is formed at the anode when copper sulfate undergoes electrolysis using inert electrodes is oxygen.

Three factors that influence the species formed at the electrode are:

• The relative values of E^θ

• The concentration of the ions present

• The identity of the electrode

Copper is formed at the cathode as copper ions are preferentially reduced over hydrogen as has a more positive  E^θ value.

Oxygen is formed at the anode during the electrolysis of sodium chloride with a low concentration.

Chlorine is formed at the anode during the electrolysis of sodium chloride with a high concentration.

Electrodes that take part in the redox processes are known as active electrodes.

Inert electrodes such as platinum and carbon are called passive electrodes.

The half- equation for the formation of copper ions at the anode during the purification of copper is:

Cu (s)  → Cu²⁺ (aq) + 2e^- 

Electroplating is the process of depositing a thin layer of metal onto an object's surface using an electric current.

Two reasons why electroplating of metals is carried out include:

• Corrosion protection/ resistance

• To give the metal a shiny appearance

• To improve the electrical conductivity

• To improve the strength

The following reaction takes place at the cathode as it shows reduction.

[Ag(CN)₂]^- (aq) + e^-  →  Ag (s) + 2CN^- (aq) 

The purpose of the anode in electroplating is to replenish the loss of the metal during electrolysis and maintain a constant concentration of the electrolyte.

Two factors that influence the amount of metal deposited during electroplating include:

• Current

• Time

• The size or surface area of the electrode

• The temperature of the solution

• The concentration of the metal electrolyte

True.

The cathode will increase in mass during electroplating.

True.

The anode will decrease in mass during electroplating.