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
what happens if current is allowed to flow
reactions then occur separately at each electrode
voltage will fall to zero and reactants are used up
most positive will always undergo reduction
most negative will always undergo oxidation
what are standard electrode potentials always measured as
reduction values
most positive will always undergo reduction
most negative will always undergo oxidation
measuring the electrode potential of the cell
can’t measure the absolute potential of a half electrode on its own but can measure the potential difference between the two half-cells measured
connect to another half cell of a known potential and calculate difference
connect to standard hydrogen cell
what is the reference electrode we use
hydrogen electrode
potential of 0 volts
what is the standard hydrogen equilibrium
H2 (g) (reversible) 2H+ (aq) + 2e-
components of a standard hydrogen electrode
- hydrogen gas at pressure 100kPa
- solution containing the hydrogen ion at 1moldm-3
- temperature at 298K
draw the standard reference electrode
draw the two beakers
platinum as 1 electrode, 1M solution of HCl
hydrogen gas is collected at 100kPa
KNO3 salt bridge
metal electrode and metal ion solution at 1M
voltmeter
secondary standards
calibrated against the standard hydrogen cell
standard electrode that has been calibrated against the primary standard
common are:
silver/silver chloride E0= +0.22V
calomel electrode E0= +0.27V
standard electrode potential
when an electrode system is connected to the hydrogen electrode system and standard conditions apply, the potential difference measured is called the standard electrode potential
what are the standard conditions for electrode potentials
all ion solutions at 1M
temperature 298K
gases at 100kPa
no current flowing
what is the most useful application of electrode potentials
to show direction of spontaneous change for redox reactions
how do you work out the E value of a cell
E reduction- E oxidation
subtract the most positive value from the most negative value
a spontaneous change will always have a positive E value
more positive the E0
greater tendency for the species on the left to reduce and act as an oxidsing agent
more negative the E0
greater tendency for the species on the right to oxidise and act as reducing agents
most powerful reducing agents
found at most negative end of the series on the right
one with lower oxidation number
most powerful oxidising agents
found at the most positive end of the series on the left
one with higher oxidation number
what is cell EMF
measure of how far from equilibrium the cell reaction lies
more positive the more likely the reaction is to occur
effects of changing conditions on cell can be explained by what
Le Chatelier’s principle
effect of concentration on cell emf
increasing concentration of “reactants” would increase emf and decreasing them would cause emf to decrease
effect of temperature on cell emf
most cells are exothermic in the spontaneous direction so this would result in a decrease in E cell
even if E value is positive what may not occur
the reaction
may occur so slowly or not occur at all
if reaction has a high activation energy it’ll not occur
explaining homogeneous catalysis using E values
reaction between I- and S2O82- is catalysed by Fe2+
uncatalyzed is very slow as reaction needs a collision between 2 negative ions, but the repulsion results in a high activation energy
for a substance to act as a homogeneous catalyst its electrode potential value must lie between the electrode potentials of the 2 reactants so it can oxidise one and reduce the other
cells
electrochemical cells can be used as a commercial source of electrical energy
can be non-rechargeable, rechargeable and fuel cells
when are cells non-rechargeable
when reactions that occur within them are non-reversible
fuel cells
use the energy from the reaction of a fuel with oxygen to create a voltage
what are fuel cells vehicles fuelled by
hydrogen gas
hydrogen-rich fuels
hydrogen fuel cell
uses a potassium hydroxide electrolyte
overall reaction: 2H2 + O2 -> 2H2O
voltage of fuel cells
remains constant as they are continuously fed with fresh O2 and H2 so they maintain a constant conc of reactants
ordinary cells the voltage drops over time as the reactant conc drops
fuel cells in acidic conditions
e cell is the same as alkaline conditions as the overall equation is the same
using standard conditions fuel cells
rate is too slow to produce an appreciable current
higher temps are therefore used to increase the rate but reaction is then exothermic so emf falls
higher pressure counteracts this
advantages of fuel cells over conventional petrol or diesel-powered vehicles
less pollution and CO2 (pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2)
greater efficiency
limitations of hydrogen fuel cells
storing and transporting hydrogen in terms of safety, feasibility of a pressurised liquid and a limited life cycle of a solid “adsorber” or “absorber”
limited lifetime (regular replacement) high production costs
use of toxic chemicals in their production
