5.2.3 Electrode potential and fuel cells

Question 1

oxidation

Answer 1

loss of electrons

increase in oxidation number

Question 2

reduction

Answer 2

gain of electrons

decrease in oxidation number

Question 3

in an oxidation half equation where are the electrons

Answer 3

on the right

Question 4

in a reduction half equation where are the electrons

Answer 4

on the left

Question 5

reducing agent

Answer 5

electron donors

get oxidised

Question 6

oxidising agents

Answer 6

electron acceptors

get reduced

Question 7

how do you combine half equations

Answer 7

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

Question 8

common redox titrations

Answer 8

thiosulfate

manganate

Question 9

thiosulfate redox reaction

Answer 9

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

Question 10

manganate redox titration

Answer 10

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

Question 11

choosing the correct acid for a manganate titration

Answer 11

need to supply 8H+ ions
use dilute sulfuric acid
in excess

Question 12

why can’t you use ethanoic acid in manganate titration

Answer 12

can’t supply the large amount of H+ ions needed

Question 13

why can’t you use conc HCl in manganate titrations

Answer 13

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

Question 14

why can’t nitric acid be used in manganate titration

Answer 14

as it is an oxidising agent
will oxidise Fe2+ to Fe3+
as E0 of NO3 is greater than the fe2 and fe3

Question 15

other useful manganate titrations

Answer 15

with hydrogen peroxide 
with ethanedioate (slower so heat to 60 degrees)

Question 16

electrochemical cells

Answer 16

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

Question 17

drawing the half-cells

Answer 17

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

Question 18

why does a voltage form in half-cells

Answer 18

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

Question 19

why use a high resistance voltmeter

Answer 19

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

Question 20

salt bridge

Answer 20

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

Question 21

why isn’t a wire used as a salt bridge

Answer 21

wire isn’t used as it would establish its own electrode system with the solutions

Question 22

what happens if current is allowed to flow

Answer 22

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

Question 23

what are standard electrode potentials always measured as

Answer 23

reduction values
most positive will always undergo reduction
most negative will always undergo oxidation

Question 24

measuring the electrode potential of the cell

Answer 24

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

Question 25

what is the reference electrode we use

Answer 25

hydrogen electrode | potential of 0 volts

Question 26

what is the standard hydrogen equilibrium

Answer 26

H2 (g) (reversible) 2H+ (aq) + 2e-

Question 27

components of a standard hydrogen electrode

Answer 27

1. hydrogen gas at pressure 100kPa 2. solution containing the hydrogen ion at 1moldm-3 3. temperature at 298K

Question 28

draw the standard reference electrode

Answer 28

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

Question 29

secondary standards

Answer 29

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

Question 30

standard electrode potential

Answer 30

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

Question 31

what are the standard conditions for electrode potentials

Answer 31

all ion solutions at 1M temperature 298K gases at 100kPa no current flowing

Question 32

what is the most useful application of electrode potentials

Answer 32

to show direction of spontaneous change for redox reactions

Question 33

how do you work out the E value of a cell

Answer 33

E reduction- E oxidation subtract the most positive value from the most negative value a spontaneous change will always have a positive E value

Question 34

more positive the E0

Answer 34

greater tendency for the species on the left to reduce and act as an oxidsing agent

Question 35

more negative the E0

Answer 35

greater tendency for the species on the right to oxidise and act as reducing agents

Question 36

most powerful reducing agents

Answer 36

found at most negative end of the series on the right | one with lower oxidation number

Question 37

most powerful oxidising agents

Answer 37

found at the most positive end of the series on the left | one with higher oxidation number

Question 38

what is cell EMF

Answer 38

measure of how far from equilibrium the cell reaction lies | more positive the more likely the reaction is to occur

Question 39

effects of changing conditions on cell can be explained by what

Answer 39

Le Chatelier's principle

Question 40

effect of concentration on cell emf

Answer 40

increasing concentration of "reactants" would increase emf and decreasing them would cause emf to decrease

Question 41

effect of temperature on cell emf

Answer 41

most cells are exothermic in the spontaneous direction so this would result in a decrease in E cell

Question 42

even if E value is positive what may not occur

Answer 42

the reaction may occur so slowly or not occur at all if reaction has a high activation energy it'll not occur

Question 43

explaining homogeneous catalysis using E values

Answer 43

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

Question 44

cells

Answer 44

electrochemical cells can be used as a commercial source of electrical energy can be non-rechargeable, rechargeable and fuel cells

Question 45

when are cells non-rechargeable

Answer 45

when reactions that occur within them are non-reversible

Question 46

fuel cells

Answer 46

use the energy from the reaction of a fuel with oxygen to create a voltage

Question 47

what are fuel cells vehicles fuelled by

Answer 47

hydrogen gas | hydrogen-rich fuels

Question 48

hydrogen fuel cell

Answer 48

uses a potassium hydroxide electrolyte | overall reaction: 2H2 + O2 -> 2H2O

Question 49

voltage of fuel cells

Answer 49

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

Question 50

fuel cells in acidic conditions

Answer 50

e cell is the same as alkaline conditions as the overall equation is the same

Question 51

using standard conditions fuel cells

Answer 51

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

Question 52

advantages of fuel cells over conventional petrol or diesel-powered vehicles

Answer 52

less pollution and CO2 (pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2) greater efficiency

Question 53

limitations of hydrogen fuel cells

Answer 53

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