Redox Flashcards by TAN ZHENGYUE HCI

Define a redox reaction.

A redox reaction is a chemical reaction involving both oxidation and reduction simultaneously.

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Define oxidation, in terms of oxygen / hydrogen gain / loss.

A process whereby a substance gains oxygen or loses hydrogen in a chemical reaction.

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Define reduction, in terms of oxygen / hydrogen gain / loss.

A process whereby a substance loses oxygen or gains hydrogen in a chemical reaction.

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Define the oxidation number of an atom.

The charge that the atom carries as a monoatomic ion.

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In a compound, state the nature of the oxidation number of the more electronegative element.

The oxidation number of the more electronegative element is negative.

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State the oxidation number (O.S.) for the following elements as well as the exceptions, if any.
1: hydrogen, H
2: oxygen, O
3: Group 1 metals
4: Group 2 metals
5: aluminium, Al
6: fluorine, F
7: chlorine, Cl

1: +1
Exception: O.S. of H in metal hydride is -1
2: -2
Exception: O.S. of O in peroxide is -1
3: +1
4: +2
5: +3
6: -1
7: -1

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Why does fluorine have an oxidation number of -1 in all compounds?

Fluorine is the most electronegative element. During ionic bonding, flouoine takes oxidation number -1 due to the fluoride anion. During covalent bonding, fluorine forms one covalent bond. Hence, the shared electron pair in the single covalent bond will be assigned to fluorine and it takes oxidation number of -1.

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State the oxidation number (O.S.) of each element in the following compounds.
1: MgBr2
2: CO2
3: Cl2
4: H2O2
5: SF4
6: SO3

1:
O.S. of Mg= +2
O.S. of Br= -1
2:
O.S. of C= +4
O.S. of O= -2
3:
O.S. of Cl= 0, since there is no difference in electronegativity between the 2 Cl atoms
4:
O.S. of H= +1
O.S. of O= -1
5:
O.S. of S= +4
O.S. of F= -1
6:
O.S. of S= +6
O.S. of O= -2

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Define oxidation, in terms of electron transfer.

Oxidation is a process whereby a substance loses electrons, causing an increase in oxidation number.

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Define reduction, in terms of electron transfer.

Reduction is a process whereby a substance gains electrons, causing a decrease in oxidation number.

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In terms of changes in oxidation number, state which substance is oxidised and reduced.
2NH3 (g) + 3CuO (s) -> N2 (g) + 3Cu (s) + 3H2O (l)

NH3 is oxidised as there is an increase in oxidation number of N from -3 in NH3 to 0 in N2.
CuO is reduced as there is a decrease in oxidation number of Cu from +2 in CuO to 0 in Cu.
(note the format of answer)

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Identify the oxidising agents and reducing agents in the following substances:
1. oxygen, O2
2. carbon, C
3. hydrogen, H2
4. chlorine, Cl2
5. carbon monoxide, CO
6. aluminium, Al
7. zinc, Zn
8. potassium manganate (VII), KMnO4
9. hydrogen peroxide, H2O2
10. potassium iodide, KI
11. potassium dichromate (VI), K2Cr2O7

Oxidising agents:
1, 4, 8, 9, 11

Reducing agents:
2, 3, 5, 6, 7, 10

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For the redox reactions below, identify the reducing and oxidising agents and explain why.
Reaction 1:
H2O2 (aq) + 2Fe^2+ (aq) + 2H^+ (aq) → 2Fe^3+ (aq) + 2H2O (l)

Reaction 2:
NO3^− (aq) + 4Zn (s) + 7OH^– (aq) + 6H2O (l) → 4Zn(OH)4^2– (aq) + NH3 (aq)

Reaction 1:
Fe^2+ is the reducing agent as it reduces H2O2 since there is a decrease in oxidation number of O from -1 in H2O2 to -2 in H2O.
H2O2 is the oxidising agent as it oxidises Fe^2+ since there is an increase in oxidation number of Fe from +2 in Fe^2+ to +3 in Fe^3+.

Reaction 2:
Zn is the reducing agent as it reduces NO3^− since there is a decrease in oxidation
number of N from +5 in NO3^− to −3 in NH3.
NO3^− is the oxidising agent as it oxidises Zn since there is an increase in oxidation number of Zn from 0 in Zn to +2 in Zn(OH)4^2−.

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Which substance can act as both oxidising agent and reducing agent?

hydrogen peroxide, H2O2

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In general, metals are reducing agents and halogens are oxidising agents.

True/False?

True

The more reactive the metal, the more powerful it is as a reducing agent, since it would give electrons away readily.

In general, halogens are oxidising agents. E.g. Cl2 accepts electrons readily to form 2Cl^-.

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State the reducing agent used to test for oxidising agent.

aqueous potassium iodide, KI
(the K+ ion is the spectator ion as I- ion is the species that causes reduction of another substance)

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Describe the procedure of testing for oxidising agent using aqueous potassium iodide, KI, and state the positive observation. Write the half equation.

Procedure: Add a few drops of aqueous potassium iodide to the solution to be tested.

Positive observation: Colourless solution (containing KI) turns brown. This is because I- ion is colourless and during the reaction, it is oxidised to I2 which is brown.

Half equation: 2I^- (aq) -> I2(aq) + 2e^-

State the colour change of Fe^2+ when it is used to test for oxidising agents.

State the colour of precipitate formed on adding aqueous NaOH or ammonia afterwards.

Pale green Fe^2+ turns yellow Fe^3+.

Reddish-brown precipitate formed on adding aqueous NaOH or ammonia.

State the colour change of Fe^3+ when it is used to test for reducing agents.

State the colour of precipitate formed on adding aqueous NaOH or ammonia afterwards.

Yellow Fe^3+ turns pale green Fe^2+.

Green precipitate formed on adding aqueous NaOH or ammonia.

Write the chemical equation, with state symbols, of the oxidation of hydrogen peroxide.

H2O2 (aq) -> O2 (g) + 2H^+ (aq) + 2e^-

Write the chemical equation, with state symbols, of the oxidation of sulfur dioxide.

SO2 (g) + 2H2O (l) -> SO4^2- (aq) + 4H^+ (aq) + 2e^-

State the observation of the oxidation of hydrogen sulfide, H2S.
Hence, write the chemical equation of the oxidation of hydrogen sulfide.

Pale yellow precipitate of sulfur is deposited during the reaction.

H2S -> S + 2H^+ + 2e^-

List the oxidising agents that can be used to test for reducing agent.

acidified potassium manganate(VII), KMnO4

(the K+ ion is the spectator ion as MnO4- ion is the species that causes oxidation of another substance)

acidified potassium dichromate(VI), K2Cr2O7

(the K+ ion is the spectator ion as Cr2O7 ^2- ion is the species that causes oxidation of another substance)

Describe the procedure of testing for reducing agent using aqueous potassium manganate(VII), KMnO4, and state the positive observation. Write the half equation.

Procedure: Add dilute sulfuric acid, followed by a few drops of aqueous potassium manganate.

Positive observation: Purple solution (containing KMnO4) turns colourless. This is because MnO4^- ion is purple in colour and during the reaction, it is reduced to Mn^2+ which is colourless.

Half equation: MnO4^- (aq) + 8H^+ (aq) + 5e^- -> Mn^2+ (aq) + 4H2O (l)

SO2 gas is identified by using a piece of filter paper soaked with acidified potassium manganate(VII) at the mouth of a test tube. If SO2 is present, the filter paper turns from purple to colourless. Determine if SO2 is an oxidising or reducing agent.

Reducing agent.

Describe the procedure of testing for reducing agent using aqueous potassium dichromate(VI), K2Cr2O7, and state the positive observation. Write the half equation.

Add dilute sulfuric acid followed by aqueous potassium dichromate(VI). Positive observation: Orange solution turns green. This is because Cr2O7^2- ion is orange in colour and during the reaction, it is reduced to Cr ^3+ which is green. Half equation: Cr2O7^2- (aq) + 14H^+ (aq) + 6e^- -> 2Cr^3+ (aq) + 7H2O (l)

Describe the colour change that will be observed when potassium iodide is gradually added to a solution containing acidified potassium manganate(VI).

Purple solution turns brown.

Given that Fe is above Cu in the reactivity series, write the end products (with state symbols) of the reaction below. Fe(s) + CuSO4(aq)

FeSO4(aq) and Cu(s)

A more reactive metal loses electrons and become oxidised while a less reactive metal gains electron and become reduced. True / False?

True

Reactions of three metals, W, X and Y, or compounds of these metals, are described. 1. The oxide of W is reduced both by heating with carbon and by heating with hydrogen. 2. The oxide of X is not reduced by heating with carbon or by heating with hydrogen. 3. W displaced Y from an aqueous solution of the sulfate of Y. List the three metals from least reactive to most reactive.

Y, W, X

Define disproportionation.

Disproportionation is a redox reaction in which a single substance is simultaneously oxidised and reduced to form two different products.

Identify the disproportionation reactions from the following reactions. Reaction 1: H2SO4 + 2HBr → SO2 + Br2 + 2H2O Reaction 2: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O Reaction 3: IO3 ^- + 5I ^- + 6H ^+ -> 3I2 + 3H2O Reaction 4: 2NO2 + H2O → HNO3 + HNO2 Reaction 5: 2H2O2 → 2H2O + O2

Reactions 2, 4, 5

Why is an external indicator not required to be added in potassium manganate(VII) titration?

Since KMnO4(aq) is purple in colour while its reduced product, Mn ^2+(aq), is faint pink which is essentially colourless, an external indicator is not required to be added in the course of the titration. The endpoint is taken when the first permanent pink coloration is formed because of an excess drop of potassium manganate(VII) added.

State the half equation of the reduction of the manganate(VII) ion, MnO4.

MnO4^- (aq) + 8H^+ (aq) + 5e^- -> Mn^2+ (aq) + 4H2O (l)

Why is an external indicator not required to be added in potassium dichromate(VI) titration?

Upon reduction in an acidic medium, the orange dichromate(VI) ion changes to green chromium(III) ions.

State the half equation of the reduction of the dichromate(VI) ion, Cr2O7.

Cr2O7^2- (aq) + 14H^+ (aq) + 6e^- -> 2Cr^3+ (aq) + 7H2O (l)

State the 2 main substances involved in an iodometric titration and briefly state its function.

Thiosulfate ions and iodine Function: To determine the amount of iodine present in a solution.

Write the chemical equation, with state symbols, of the reaction of iodine and thiosufate ions, S2O3^2-.

I2 (aq) + 2S2O3^2- (aq) -> 2I^- (aq) + S4O6^2- (aq) (in S4O6, 2 of the sulfur atoms present in the ion are in oxidation state of 0 and the other 2 atoms are in oxidation state of +5)

Describe an iodometric titration as well as the endpoint of the titration.

As thiosulfate ions are added, it would reduce the iodine, resulting in a decrease in the concentration of iodine. The resultant solution will approach a pale yellow coloration. The endpoint of the titration occurs when a distinct colour change from deep blue to colourless is observed.

Suggest what can be done to obtain a more accurate detection of the endpoint during an iodometric titration.

Add a few drops of starch solution when the solution turns pale yellow.

Balance the redox reaction below using the half-equation method. I2 + S2O3^2- -> I^- + S4O6^2-

[R] I2 + 2e^- -> 2I^- [O] 2S2O3^2- -> S4O6^2- + 2e^- Overall: I2 (aq) + 2S2O3^2- (aq) -> 2I^- (aq) + S4O6^2- (aq)