Redox reactions worksheet for chemistry students.

Worksheet: Redox Reactions

Let's solve the problems step by step.

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#### 1. Define the process of oxidation in terms of:

Oxidation is the process where a substance loses electrons, gains oxygen, or loses hydrogen. Here are the definitions for each part:

- a) Transfer (gain or loss) of oxygen:
Oxidation involves the gain of oxygen by a substance.

- b) Transfer (gain or loss) of electrons:
Oxidation involves the loss of electrons by a substance.

- c) Transfer (gain or loss) of hydrogen:
Oxidation involves the loss of hydrogen by a substance.

- d) Decrease or increase in oxidation number:
Oxidation involves an increase in the oxidation number of an atom.

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#### 2. Define the process of reduction in terms of:

Reduction is the process where a substance gains electrons, loses oxygen, or gains hydrogen. Here are the definitions for each part:

- a) Transfer (gain or loss) of oxygen:
Reduction involves the loss of oxygen by a substance.

- b) Transfer (gain or loss) of electrons:
Reduction involves the gain of electrons by a substance.

- c) Transfer (gain or loss) of hydrogen:
Reduction involves the gain of hydrogen by a substance.

- d) Decrease or increase in oxidation number:
Reduction involves a decrease in the oxidation number of an atom.

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#### 3. Complete the half equations by balancing them and identify each half reaction as either an oxidation (O) or reduction (R) reaction.

We will balance each equation and determine whether it represents oxidation or reduction based on electron transfer.

##### (i) Na(s) → Na⁺(aq)
- Sodium loses one electron to form Na⁺.
- Balanced equation:
\[
\text{Na}(s) \rightarrow \text{Na}^+(aq) + e^-
\]
- This is an oxidation (O) reaction because sodium loses electrons.

##### (ii) Mg²⁺(aq) → Mg(s)
- Magnesium gains two electrons to form Mg.
- Balanced equation:
\[
\text{Mg}^{2+}(aq) + 2e^- \rightarrow \text{Mg}(s)
\]
- This is a reduction (R) reaction because magnesium gains electrons.

##### (iii) Fe(s) → Fe³⁺(aq)
- Iron loses three electrons to form Fe³⁺.
- Balanced equation:
\[
\text{Fe}(s) \rightarrow \text{Fe}^{3+}(aq) + 3e^-
\]
- This is an oxidation (O) reaction because iron loses electrons.

##### (iv) Cl⁻(aq) → Cl₂(g)
- Chloride ion loses one electron per chlorine atom to form Cl₂.
- Balanced equation:
\[
2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-
\]
- This is an oxidation (O) reaction because chloride loses electrons.

##### (v) K(s) → K⁺(aq)
- Potassium loses one electron to form K⁺.
- Balanced equation:
\[
\text{K}(s) \rightarrow \text{K}^+(aq) + e^-
\]
- This is an oxidation (O) reaction because potassium loses electrons.

##### (vi) O₂(g) → O²⁻(aq)
- Oxygen gains four electrons to form two O²⁻ ions.
- Balanced equation:
\[
\text{O}_2(g) + 4e^- \rightarrow 2\text{O}^{2-}(aq)
\]
- This is a reduction (R) reaction because oxygen gains electrons.

##### (vii) F⁻(aq) → F₂(g)
- Fluoride ion loses one electron per fluorine atom to form F₂.
- Balanced equation:
\[
2\text{F}^-(aq) \rightarrow \text{F}_2(g) + 2e^-
\]
- This is an oxidation (O) reaction because fluoride loses electrons.

##### (viii) Cu²⁺(aq) → Cu(s)
- Copper gains two electrons to form Cu.
- Balanced equation:
\[
\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)
\]
- This is a reduction (R) reaction because copper gains electrons.

##### (ix) S(s) → S²⁻(aq)
- Sulfur gains two electrons to form S²⁻.
- Balanced equation:
\[
\text{S}(s) + 2e^- \rightarrow \text{S}^{2-}(aq)
\]
- This is a reduction (R) reaction because sulfur gains electrons.

##### (x) H₂(g) → H⁺(aq)
- Hydrogen loses one electron per hydrogen atom to form H⁺.
- Balanced equation:
\[
\text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^-
\]
- This is an oxidation (O) reaction because hydrogen loses electrons.

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#### 4. Consider the incomplete reaction (partial ionic equation) given below:
\[ \text{Cu}^{2+} + \text{Zn} \rightarrow \text{Zn}^{2+} + \text{Cu} \]

##### (i) Give the two half equations for this reaction: oxidation + reduction

1. Oxidation half-reaction (Zn → Zn²⁺):
Zinc loses two electrons to form Zn²⁺.
\[
\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-
\]

2. Reduction half-reaction (Cu²⁺ → Cu):
Copper(II) ion gains two electrons to form Cu.
\[
\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)
\]

##### (ii) Identify the chemical substance (species) that has undergone oxidation

- The species that undergoes oxidation is Zn(s) because it loses electrons.

##### (iii) Identify the chemical substance (species) that has undergone reduction

- The species that undergoes reduction is Cu²⁺(aq) because it gains electrons.

##### (iv) Identify (from the reactants) the oxidising agent (oxidant) in this reaction

- The oxidising agent is the species that causes oxidation by accepting electrons. In this case, it is Cu²⁺(aq) because it accepts electrons from Zn.

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Final Answers:

1.
- a) Gain of oxygen
- b) Loss of electrons
- c) Loss of hydrogen
- d) Increase in oxidation number

2.
- a) Loss of oxygen
- b) Gain of electrons
- c) Gain of hydrogen
- d) Decrease in oxidation number

3.
- (i) \(\text{Na}(s) \rightarrow \text{Na}^+(aq) + e^-\) (O)
- (ii) \(\text{Mg}^{2+}(aq) + 2e^- \rightarrow \text{Mg}(s)\) (R)
- (iii) \(\text{Fe}(s) \rightarrow \text{Fe}^{3+}(aq) + 3e^-\) (O)
- (iv) \(2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-\) (O)
- (v) \(\text{K}(s) \rightarrow \text{K}^+(aq) + e^-\) (O)
- (vi) \(\text{O}_2(g) + 4e^- \rightarrow 2\text{O}^{2-}(aq)\) (R)
- (vii) \(2\text{F}^-(aq) \rightarrow \text{F}_2(g) + 2e^-\) (O)
- (viii) \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\) (R)
- (ix) \(\text{S}(s) + 2e^- \rightarrow \text{S}^{2-}(aq)\) (R)
- (x) \(\text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^-\) (O)

4.
- (i) Oxidation: \(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-\)
Reduction: \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\)
- (ii) \(\text{Zn}(s)\)
- (iii) \(\text{Cu}^{2+}(aq)\)
- (iv) \(\text{Cu}^{2+}(aq)\)

\boxed{\text{See detailed explanations above}}