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Topic 9: Redox Processes

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

oxidation

Back (Answer)

• loss of electrons • increase in oxidation no • gain of oxygen • loss of hydrogen

#2
Front (Question)

reduction

Back (Answer)

• gain of electrons • decrease in oxidation no • loss of oxygen • gain of hydrogen

#3
Front (Question)

oxidation state

Back (Answer)

apparent charge of an atom in a molecule or ion. • used to measure electron control or possession relative to the atom in pure element

#4
Front (Question)

relation between reducing power and reactivity in metals

Back (Answer)

more reactive metals are stronger reductants

#5
Front (Question)

relation between oxidizing power and reactivity in non-metals

Back (Answer)

more reactive non-metals are stronger oxidizing agents

#6
Front (Question)

redox titration

Back (Answer)

redox reaction between an oxidizing agent and reducing agent

#7
Front (Question)

chemically speaking, what happens in a redox titration?

Back (Answer)

electrons are transferred from reductant to oxidant

#8
Front (Question)

redox titrations: analysis of iron with manganate (VII)

Back (Answer)

5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O • uses KMnO4 in acidic soln. as oxidant • oxidises Fe (II) ions to Fe (III) • MnO4- is reduced to Mn2+ • colour change: deep purple to colourless

#9
Front (Question)

redox titrations: iodine-thiosulfate reaction

Back (Answer)

2I- (aq) + oxidant -> I2 (aq) + reduced product • oxidant reacts with excess iodides to form iodine diatomic molecules • oxidants can be KMnO4, KIO3, K2Cr2O7, NaOCl, etc. 2S2O3 2- (aq) + I2 (aq) -> 2I- (aq) + S4O6 2- (aq) • the liberated iodine is then titrated with sodium thiosulfate (Na2S2O3) • using starch as an indicator (NOT added at the start but during titration • initially forms deep blue colour due to starch, but as I2 is reduced to I-, the blue colour disappears

#10
Front (Question)

redox titrations: winkler method (what it is, its function, etc)

Back (Answer)

• calculates dissolved oxygen content of water • used to measure degree of pollution • as oxygen is used by bacteria in decomposition reactions

#11
Front (Question)

biological oxygen demand (BOD)

Back (Answer)

• amount of oxygen used to decompose organic matter in a sample of water over a specified time period • usually 5 days at a specified temp • high BOD = high quantity of degradable organic waste = low level of dissolved oxygen

#12
Front (Question)

redox titrations: winkler method (process)

Back (Answer)

• Dissolved oxygen O2 (g) is fixed by the addition of a manganese (II) salt, such as MnSO4. This causes oxidation of Mn (II) to higher oxidation states: 2Mn2+ (aq) + O2 (g) + 4OH- (aq) -> 2MnO2 (s) + 2H2O (l) • Acidified iodide ions (I-) are added and are oxidised by Mn (IV) to I2: MnO2 (s) + 2I- (aq) + 4H+ (aq) -> Mn2+ (aq) + I2 (aq) + 2H2O (l) • Iodine produced is titrated with sodium thiosulfate: 2S2O3 2- (aq) + I2 (aq) -> 2I- (aq) + S4O6 2- (aq)

#13
Front (Question)

redox titrations: winkler method (ratio of O2 : S2O3 2-)

Back (Answer)

1:4

#14
Front (Question)

voltaic cells

Back (Answer)

• generates electricity from spontaneous redox reactions • separates two half-reactions into half-cells, allowing e-s to flow between them through an external circuit 2 connected half-cells = 1 voltaic cell

#15
Front (Question)

electrode potential

Back (Answer)

charge separation between the metal and its ions in solution

#16
Front (Question)

half cell

Back (Answer)

• where a half-reaction occurs - simplest one is made by putting a strip of metal into a solution of its ions

#17
Front (Question)

relationship between cell equilibrium and reactivity (of metal)

Back (Answer)

• more reactive metals are stronger reducing agents • they have a higher tendency to lose electrons • the less reactive the metal, the more to the right the equilibrium position will be

#18
Front (Question)

voltaic cell connections between half-cells

Back (Answer)

• external electronic circuit - salt bridge

#19
Front (Question)

external electronic circuit

Back (Answer)

• connected to the metal electrode in each half-cell • can have a voltmeter attached to record generated voltage • electrons flow from anode to cathode through wire

#20
Front (Question)

salt bridge

Back (Answer)

• glass tube/strip of absorptive paper containing an aq soln of ions • movement of these ions neutralise build-up of charge and maintains potential difference • anions move in salt bridge from cathode to anode, which opposes the flow of e-s in external circuit • cations move in salt bridge from anode to cathode • solution chosen is typically aq NaNO3 or KNO3

#21
Front (Question)

what is affected by the difference in reducing strength between electrodes in a voltaic cell?

Back (Answer)

• direction of electron flow (e-s always flow to the electrode with less reducing power) • voltage generated (the greater the reducing power difference, the greater the voltage generated)

#22
Front (Question)

electromotive force

Back (Answer)

• potential difference • generated due to e- flow between half-cells • depends on difference of reducing power • also called cell potential/electrode potential

#23
Front (Question)

standard hydrogen electrode

Back (Answer)

reference standard for measuring reducing power of half-cells

#24
Front (Question)

standard hydrogen electrode

Back (Answer)

• modified form of pH electrode • platinum is used as the conducting metal in the electrode • as platinum is inert and doesn’t ionise • it also acts as a catalyst for proton reduction reaction • the form of platinum used is platinized platinum (the surface of the metal is covered with very finely divided platinum, or platinum black) • this causes the electrode reaction to occur rapidly • due to the large surface area helping in the adsorption of hydrogen gas