5.2.3 Electrode potential and fuel cells Flashcards
oxidation
loss of electrons
increase in oxidation number
reduction
gain of electrons
decrease in oxidation number
in an oxidation half equation where are the electrons
on the right
in a reduction half equation where are the electrons
on the left
reducing agent
electron donors
get oxidised
oxidising agents
electron acceptors
get reduced
how do you combine half equations
multiply half equations to get equal electrons
add half equations together and cancel electrons
multiply half equations to get equal electrons
add half equations together and cancel electrons
common redox titrations
thiosulfate
manganate
thiosulfate redox reaction
2(S2O3)^2- (aq) + I2 (aq) -> 2I- (aq) +S4O6 2- (aq)
yellow/brown to colourless
starch indicator is added near the end point when iodine fades a pale yellow, colour change is blue/black to colourless
manganate redox titration
self-indicating
MnO4- (aq) + 8H+ (aq) + 5Fe 2+ (aq) -> Mn 2+ (aq) + 4H2O (l) + 5Fe 3+ (aq)
purple to colourless
if manganate is in the burette then colourless to purple will be change
choosing the correct acid for a manganate titration
need to supply 8H+ ions
use dilute sulfuric acid
in excess
why can’t you use ethanoic acid in manganate titration
can’t supply the large amount of H+ ions needed
why can’t you use conc HCl in manganate titrations
as Cl- ions would be oxidised to Cl2 by MnO4-
as E0 value of Mn is greater than the Cl
leads to greater amount of manganate being used and toxic Cl2 gas being produced
why can’t nitric acid be used in manganate titration
as it is an oxidising agent
will oxidise Fe2+ to Fe3+
as E0 of NO3 is greater than the fe2 and fe3
other useful manganate titrations
with hydrogen peroxide with ethanedioate (slower so heat to 60 degrees)
electrochemical cells
a cell contains 2 half-cells
these are connected via a salt bridge
simple half cells contain a metal (acts as an electrode) and a 1.0 moldm-3 solution containing that metal
these will produce a small voltage if connected into a circuit
drawing the half-cells
2 beakers
1 with 1 metal electrode and a 1M solution of the metal
the other with another metal but exactly the same
connect them up through a voltmeter
electron flow needs to be drawn in the correct direction
MUST DRAW AND LABEL SALT BRIDGE BETWEEN BEAKERS
why does a voltage form in half-cells
when connected, one of the metals will have a greater tendency to oxidise than the other
place the one with the most positive E0 value on the bottom and follow the anti-clockwise rule
more electrons will build up on one electrode than the other so potential difference is established
why use a high resistance voltmeter
to stop the current flowing in a circuit
to measure the maximum possible E value
reactions will not be occurring as voltmeter stops current from flowing
salt bridge
used to connect the circuit
free moving ions conduct the charge
made of filter paper soaked in salt solution e.g. potassium nitrate
salt should be unreactive with electrodes and electrode solutions
why isn’t a wire used as a salt bridge
wire isn’t used as it would establish its own electrode system with the solutions