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CHEM10007 · Fundamentals Of Chemistry

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Chapter 6 of 11 · CHEM10007

Redox & Electrochemistry

Reactions that move electrons. You assign oxidation numbers, identify oxidant and reductant, and balance redox half-equations in acidic solution. You then build and label a galvanic cell — anode, cathode, salt bridge and electron flow — and compute E°cell from standard reduction potentials, before contrasting spontaneous galvanic cells with driven electrolysis. Redox titrations are a high-yield Section B archetype.

In this chapter

What this chapter covers

  • 01Oxidation Is Loss, Reduction Is Gain of electrons (OIL RIG); oxidant is reduced, reductant is oxidised
  • 02Assigning oxidation numbers; a change in oxidation number signals a redox reaction
  • 03Balancing acidic redox: balance atoms, then O with H2O, H with H+, charge with e; combine half-equations
  • 04Galvanic (voltaic) cell: anode = oxidation, cathode = reduction; salt bridge keeps electroneutrality
  • 05Electron flow through the external wire from anode → cathode; cation migration to the cathode
  • 06cell = E°reduction + E°oxidation; E°cell > 0 means spontaneous
  • 07Electrolysis (electrolytic cell): non-spontaneous, externally driven; predicting electrode reactions
  • 08Uses of electrolysis (electroplating, metal extraction) and the galvanic-vs-electrolytic sign convention
Worked example · free

Galvanic cell from two half-cells

Q [4 marks]. A cell is built from a magnesium electrode (E° = −2.37 V) and a silver electrode (E° = +0.80 V). (a) Write the spontaneous overall reaction. (b) Calculate E°cell. (c) Identify the negative electrode and state the direction of electron flow and salt-bridge cation movement.
  • 1 mark — identifies cathode and anodeThe higher reduction potential is reduced: Ag+ + e → Ag (cathode); Mg is oxidised: Mg → Mg2+ + 2e (anode).
  • 1 mark — balanced overall reactionBalance electrons (× 2 on silver) and combine: Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s).
  • 1 mark — correct E°<sub>cell</sub>cell = E°red(cathode) − E°red(anode) = 0.80 − (−2.37) = +3.17 V; positive, so spontaneous.
  • 1 mark — cell labellingNegative electrode = Mg (the anode); electrons flow Mg → Ag through the wire; salt-bridge cations migrate toward the Ag (cathode) half-cell.
Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s); E°cell = +3.17 V; Mg is the negative anode, electrons flow Mg → Ag, and salt-bridge cations move to the silver half-cell.
Sia tip — E° values are intensive, so do NOT multiply them when you scale a half-equation — only the electron count is scaled. The species with the more positive E° is reduced at the cathode.
Glossary

Key terms

Oxidation number
A bookkeeping charge assigned to an atom by a set of rules; a change in oxidation number between reactant and product identifies oxidation or reduction.
Reductant (reducing agent)
The species that is oxidised (loses electrons) and thereby reduces another species. The oxidant is the species that is reduced.
Galvanic cell
A cell in which a spontaneous redox reaction does electrical work; oxidation occurs at the anode (negative) and reduction at the cathode (positive).
Standard cell potential (E°cell)
reduction + E°oxidation under standard conditions; a positive value indicates a spontaneous reaction.
Electrolysis
A non-spontaneous redox process driven by an external power supply in an electrolytic cell, used for electroplating and metal extraction.
FAQ

Redox & Electrochemistry FAQ

How do I balance a redox reaction in acidic solution?

Split into oxidation and reduction half-equations. In each, balance the main atoms, then balance O with H2O, balance H with H+, and balance charge with electrons. Scale so electrons lost equal electrons gained, then add the halves and cancel.

Why don't I multiply E° values when balancing electrons?

Standard reduction potentials are intensive properties — they describe a tendency, not an amount — so they are unchanged when you scale a half-equation. Only the number of electrons is multiplied to balance the overall reaction.

What is the role of the salt bridge?

It completes the circuit and maintains electroneutrality by letting ions migrate between the half-cells (anions toward the anode, cations toward the cathode) as electrons flow through the external wire. Without it the cell stops working.

Study strategy

Exam move

Lock down OIL RIG and the acidic half-equation balancing sequence, because both feed the high-value redox-titration question. Draw and label a galvanic cell from memory — anode/cathode, signs, electron and ion directions — since cell diagrams are common in Section B. Practise reading standard reduction potentials from the appendix correctly, and keep the galvanic-versus-electrolytic sign contrast clear so electrolysis questions do not trip you up.

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