5.2.3 - Redox & electrode potentials Flashcards
Constructing half equations when the oxidation no. INCREASES
Delocalised e- goes on the products side
Constructing half equations when the oxidation no. DECREASES
Delocalised e- goes on the reactants side
Half eqns. in acidic conditions
Add H2Os to balance oxygens
H+ to balance H’s
Half eqns. in alkaline conditions
Add H2Os to balance oxygens
OH- to balance H’s
Combining half eqns to get the overall eqn
Balance delocalised e- then combine the two
Delocalised e- should cancel
Redox titrations
Involves transfer of e- from one species to another
Titrations of an oxidising agent against a reducing agent
Use of acidified KMnO4
Purple potassium manganate is in the burette
Sample analysed is in the flask with an excess of dilute sulfuric acid
As the MnO4- ions react, they form Mn 2+ ions which are colourless
Why can’t HCl be used w/ MnO4-
MnO4- would oxidise Cl- to Cl2 and then affect the vol of KMnO4 required in the titration
Why can’t conc. H2SO4/HNO3 be used w/ MnO4-
They are oxidising agents themselves so affect the vol of KMnO4 required
Why can’t ethanoic acid be used w/ MnO4-
It’s a weak acid and would not provide enough H+ ions
Reacting ratio of Fe 2+ : MnO4-
5:1
How is Fe (0) used in the redox titration
Oxidised w/ H2SO4 to Fe (2+) ready for analysis
How is Fe(+3) used in the redox titration
Reacted w/ Zn to reduce it to Fe (2+)
Reacting ratio of C2O4 2-: MnO4-
2.5 : 1
C2O4 2-
Ethanedioate ion
Self indicating
Titration that does not need an indicator
Autocatalysis
A reaction where one of the products acts as a catalyst
Reaction w/ MnO4- and C2O4 2-
As both ions are -ve they repel each other and the reaction is slow and needs warming at the start
However Mn2+ acts as a catalyst and speeds up reaction
Cu^2+/ I-
2 Cu2+ + 4I- —-> 2CuI + I2
CuI - white ppt
ClO-/I-
2 I- + ClO- + 2H+ —> Cl- + I2 + H2O
I2/S2O3 2-
2 S2O3 + I2 —> 2 I- + S4O6 2-
Reacting ratio Cu2+:I2
2:1