VCE Chemistry Unit 3 AoS 2: Electrochemistry — Flashcards & Quiz

Q1: Define oxidation and reduction in terms of electron transfer.

Oxidation is the loss of electrons (increase in oxidation state). Reduction is the gain of electrons (decrease in oxidation state). They always occur together — one substance is oxidised while another is reduced (redox reaction). The substance that is oxidised is the reducing agent (it reduces the other substance). The substance that is reduced is the oxidising agent (it oxidises the other substance).

Q2: Describe the structure and operation of a galvanic cell.

A galvanic cell converts chemical energy to electrical energy using a spontaneous redox reaction. Components: 1) Two half-cells, each with an electrode in a solution of its ions. 2) External circuit — electrons flow from anode (oxidation) to cathode (reduction). 3) Salt bridge — allows ion flow to maintain electrical neutrality (anions flow toward anode, cations toward cathode). The anode is negative (electron source); cathode is positive (electron sink).

Q3: What is the standard electrode potential (E°) and how is it measured?

Standard electrode potential (E°) measures the tendency of a half-cell to be reduced relative to the standard hydrogen electrode (SHE), which is assigned E° = 0.00 V. Measured under standard conditions: 25°C, 1 mol/L solutions, 100 kPa gas pressure. More positive E° = stronger oxidising agent (greater tendency to be reduced). More negative E° = stronger reducing agent (greater tendency to be oxidised). E°cell = E°cathode - E°anode.

Q4: How is the electrochemical series used to predict redox reactions?

The electrochemical series lists half-reactions in order of decreasing E° (strongest oxidising agents at top, strongest reducing agents at bottom). To predict if a reaction is spontaneous: the oxidising agent (from the left side of a higher half-reaction) must react with the reducing agent (from the right side of a lower half-reaction). "Top-left reacts with bottom-right" = spontaneous. This gives a positive E°cell.

Q5: What is the role of the salt bridge in a galvanic cell?

The salt bridge: 1) Completes the electrical circuit by allowing ion flow between half-cells. 2) Maintains electrical neutrality — as the anode solution gains positive ions (oxidation produces cations) and the cathode solution loses positive ions (reduction consumes cations), the salt bridge provides balancing ions. 3) Contains an inert electrolyte (e.g., KNO₃ or KCl) — anions migrate toward the anode, cations toward the cathode.

Q6: Compare galvanic and electrolytic cells.

Galvanic cell: spontaneous reaction, produces electricity, ΔG < 0, anode is negative, cathode is positive. Electrolytic cell: non-spontaneous reaction, requires external electricity, ΔG > 0, anode is positive (connected to positive terminal), cathode is negative (connected to negative terminal). In BOTH: oxidation occurs at the anode, reduction occurs at the cathode. The key difference is spontaneity and energy direction.

Q7: Describe the electrolysis of water.

Water is decomposed into hydrogen and oxygen using electricity: 2H₂O(l) → 2H₂(g) + O₂(g). An electrolyte (e.g., dilute H₂SO₄ or NaOH) is added to increase conductivity. At the cathode (reduction): 4H₂O + 4e⁻ → 2H₂ + 4OH⁻ (or 4H⁺ + 4e⁻ → 2H₂ in acidic solution). At the anode (oxidation): 2H₂O → O₂ + 4H⁺ + 4e⁻. The volume of H₂ produced is twice the volume of O₂ (2:1 ratio by stoichiometry).

Q8: How is Faraday's law used to calculate amounts in electrolysis?

Faraday's law relates the amount of substance produced/consumed at an electrode to the charge passed. Key relationships: Q = It (charge = current × time), n(e⁻) = Q/F (moles of electrons = charge / Faraday constant). F = 96,485 C/mol ≈ 96,500 C/mol. Then use stoichiometry from the half-equation to find moles of substance. Finally convert to mass (m = nM) or volume (V = nVm at STP).

Q1: Oxidation involves the gain of electrons.

Answer: FALSE

Oxidation involves the LOSS of electrons (and increase in oxidation state). REDUCTION involves the gain of electrons. Remember: OIL RIG — Oxidation Is Loss, Reduction Is Gain.

Q2: In a galvanic cell, oxidation occurs at the anode.

Answer: TRUE

In ALL electrochemical cells (galvanic and electrolytic), oxidation occurs at the anode and reduction occurs at the cathode. This is by definition — AN OX, RED CAT.

Q3: A half-cell with a more positive E° value is a stronger reducing agent.

Answer: FALSE

A more positive E° indicates a stronger OXIDISING agent (greater tendency to be REDUCED). A more NEGATIVE E° indicates a stronger reducing agent (greater tendency to be oxidised).

Q4: The salt bridge in a galvanic cell allows electron flow between the two half-cells.

Answer: FALSE

The salt bridge allows ION flow (not electron flow) to maintain electrical neutrality. Electrons flow through the EXTERNAL CIRCUIT (wire) from anode to cathode. The salt bridge provides counter-ions to balance charge.

Q5: An electrolytic cell uses electrical energy to drive a non-spontaneous reaction.

Answer: TRUE

Electrolytic cells use an external power source to force a non-spontaneous reaction to occur (ΔG > 0). This is the opposite of a galvanic cell, which uses a spontaneous reaction to produce electricity.

Redox Reactions and Half-Equations

Oxidation is the loss of electrons and reduction is the gain of electrons. You must write and balance half-equations, assign oxidation numbers, and identify oxidising and reducing agents. Understanding that oxidation and reduction always occur together, and that the reducing agent is the substance oxidised, is a fundamental VCAA requirement.

Galvanic and Electrolytic Cells

Galvanic cells use spontaneous redox reactions to produce electricity, while electrolytic cells use external electricity to drive non-spontaneous reactions. You must draw and label both cell types, explain the role of the salt bridge, and remember that oxidation always occurs at the anode and reduction at the cathode in both cell types.

Electrochemical Series and Cell Voltage

The electrochemical series ranks half-reactions by standard electrode potential. You must use it to predict spontaneous reactions, calculate cell voltage using E-cell = E-cathode minus E-anode, and determine the relative strength of oxidising and reducing agents. VCAA provides the series in exams and expects confident application.

Batteries, Fuel Cells, and Corrosion

You must compare primary and secondary cells, explain how hydrogen fuel cells work, and describe the electrochemical mechanism of corrosion and its prevention methods. Understanding why sacrificial protection works even when the barrier is scratched, while tin plating accelerates corrosion at exposed areas, is a key distinction tested by VCAA.

What is the difference between galvanic and electrolytic cells in VCE Chemistry?

Galvanic cells use spontaneous redox reactions to produce electricity (E°cell > 0). Electrolytic cells use external electricity to drive non-spontaneous reactions (E°cell < 0). In both types, oxidation occurs at the anode and reduction at the cathode. The key differences are in charge labels (anode is negative in galvanic, positive in electrolytic) and energy direction.

How do you use the electrochemical series to predict reactions?

The electrochemical series lists half-reactions by E° value. To predict spontaneous reactions: the oxidising agent from a higher (more positive E°) half-reaction reacts with the reducing agent from a lower (more negative E°) half-reaction. E°cell = E°cathode - E°anode must be positive for spontaneity. Use the "top-left reacts with bottom-right" rule.

What types of batteries and fuel cells are covered in VCE Chemistry?

Key types include: primary cells (zinc-carbon, alkaline — non-rechargeable), secondary cells (lead-acid, lithium-ion — rechargeable), and hydrogen fuel cells. Students must understand how each works, compare advantages and disadvantages, and perform Faraday's law calculations for electrolysis.