redox and electrode potentials Flashcards
define oxidising agent
an oxidising agent accepts electrons and is reduced
define reducing agent
a reducing agent donates electrons and gets oxidised
define oxidation
oxidation is the loss of electrons
define reduction
reduction is a gain of electrons
what happens in a redox reaction?
- electrons are transferred from one species to another
- one element is oxidised whilst the other is reduced
write a half equation for the reaction of zinc metal with silver nitrate solution
- first write full equation (if not given) and balance it:
Zn(s) + 2AgNO3(aq) → Zn(NO3)2(aq) + 2Ag(s)
- determine the redox numbers of zinc and silverzinc: 0 → 2+ (is oxidised)silver: 1+ → 0 (is reduced)
- write half equations for each redox reaction
Zn(s) → Zn2+(aq) +2e-
2Ag+(aq) + 2e- → 2Ag (s) - balance the electron numbers in the reaction (if not balanced) and then cancel them out
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
what can you add to balance a half equation
- water (H2O)
- electrons (e-)
- H+ ions
write half equations for when acidified manganate (VII) ions (MnO4-) are reduced to Mn2+ by Fe2+ ions
- Fe2+(aq) → Fe3+ (aq) + e-
- MnO4-(s) → Mn2+ (aq)
- balance oxygen: MnO4-(aq) → Mn2+(aq) + 4H2O(l)
- balance hydrogen: MnO4-(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l)
- balance charge: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
how do you work out the concentration of a reducing agent
titrate a known volume of a reducing agent against a known concentration
describe a method that allows you to find out how many magnate ions are needed to react with a reducing agent
- measure a volume of the reducing agent using a pipette and place in a conical flask
- add dilute sulphuric acid to the flask
- gradually add MnO4- (oxidising agent) to the reducing agent using a burette and swirling the conical flask
- stop when the mixture in the flask slightly changes colour and record the volume of the oxidising agent added
- run a few titrations and calculate the mean volume of MnO4-
describe and explain the colour change in the titration using acidified potassium manganate(VII) solution
purple to colourless
- MnO4- is purple
- [Mn(H2O)6]2+ is colourless
describe and explain the colour change in the titration using acidified potassium dichromate(VI) solution
orange → green
- Cr2O7 2- is orange
- [Cr(H2O)6]3+ is violet/green
how can you make the colour change easier to spot in a titration
use a white tile underneath the conical flask
what is the iodine-sodium thiosulfate titration used for?
finding the concentration of an oxidising agent
the more concentrated an oxidising agent is the more….
the more concentrated an oxidising agent is the more ions will be oxidised by a certain volume of it
describe the method to oxidise iodide
- measure out a volume of potassium iodate (V) solution (KIO3) - the oxidising agent
- add this to an excess of acidified potassium iodide solution (KI)
- the iodate ions in the potassium iodate solution oxidise some of the iodide ions to iodine
what is the formula of sodium thiosulfate
Na2S2O3
what is the formula for thiosulfate ions
S2O3 2-
write the equation for the oxidation of iodide ions to iodine
IO3-(aq) + 5I-(aq) + 6H+(aq) → 3I2(aq) + 3H2O(l)
describe a method to titrate iodine with sodium thiosulfate
- from a burette add sodium thosulfate solution to the flask containing the oxidised iodide ions drop by drop
- when iodine colour fades to a pale yellow add 2cm of starch solution
- solution of conical flask will go dark blue to show iodine is still present
- add sodium thiosulphate one drop at a time until blue colour disappears (all iodine has been reacted)
write the equation showing the reaction between iodine with thiosulfate ions
I2(aq) + 2SO3 2-(aq) → 2I-(aq) + S4O6 2-(aq)
define a half cell
one half of an electrochemical cell constructed of a metal dipped in it’s ions, or a platinum electrode with 2 aqueous ions
when using a half cell with 2 aqueous ions what must you use?
when using a half cell with 2 aqueous ions you must use an inert but electrically conductive electrode
what happens when a rod of metal is dipped into a solution of its own ions?
an equilibrium is set up between the solid metal and the aqueous metal ions