Chapter 23

Question 1

oxidation

Answer 1

is the process of electron loss

Question 2

reduction

Answer 2

electron gain

Question 3

bromine half equation

Answer 3

Br2(aq) + 2e- -> 2 Br- (aq)

Question 4

iodine half equation

Answer 4

2I- (aq) -> I2(aq) + 2 e-

Question 5

oxidising agent

Answer 5

an electron acceptor

Question 6

reducing agent

Answer 6

an electron donor

Question 7

MnO4- + 8H+ + 5e- ->

Answer 7

Mn2+ + 4H2O

Question 8

steps for balancing equation with water and acid

Answer 8

1. balance moles
2. balance charge using electrons
3. add h2o to balance O
4. add h+ to balance h+ in h2o

Question 9

To combine two half equations there must be

Answer 9

equal numbers of electrons in the two half equations so that the electrons cancel out

Question 10

example :
Reduction MnO4- + 8 H+ + 5 e- -> Mn2+ + 4 H2O
Oxidation C2O4 2- -> 2 CO2 + 2 e-

Answer 10

multiply reduction by 2
multiply oxidation by 5
cancel electrons
overall: 2MnO4- + 16 H+ + 5C2O4 2- - > 2Mn2+ + 10 CO2 + 8 H2O

Question 11

thiosulfate Iodine redox titration

Answer 11

2S2O3 2-(aq) + I2(aq) -> 2I- (aq) + S4O6 2-(aq)

Question 12

thiosulfate redox titration colour change

Answer 12

yellow to colourless

Question 13

manganate redox titration

Answer 13

MnO4-(aq) + 8H+(aq) + 5Fe2+ (aq)-> Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

Question 14

manganate redox titration colour change

Answer 14

if iron is in the burette, purple to colourless
if manganate is in the burette, colourless to purple

Question 15

what acid is used for manganate redox titration

Answer 15

dilute sulfuric acidw

Question 16

why cant manganate redox titration be done with concentrated hcl

Answer 16

cl- would be oxidised to cl2 leading to a greater volume of mn being produced and poisonous cl2

Question 17

why cant manganate redox titration be done with nitric acid

Answer 17

nitric acid is an oxidising agent, oxidising fe2+ to fe3+ so smaller volume of mn is used

Question 18

EXAMPLE : A 2.41g nail made from an alloy containing iron is
dissolved in 100cm3 acid. The solution formed
contains Fe(II) ions.
-10cm3 portions of this solution are titrated with
potassium manganate (VII) solution of 0.02 mol dm-3
-9.80cm3 of KMnO4 were needed to react with the
solution containing the iron.
Calculate the percentage of iron by mass in the nail
MnO4-(aq) + 8H+(aq) + 5Fe2+ -> Mn2+(aq) + 4H2O + 5Fe3

Answer 18

0.02 X 9.80/1000= 1.96 X 10^-4
X 5 = 9.8 x 10^-4 mol x 10 = 9.8 x 10^-3
x 55.8 = 0.54684
0.54684/2.41 x 100 = 22.7 %

Question 19

voltaic cell

Answer 19

arranges for the 2 equations to occur separately in half cells

Question 20

electrons are transferred from one half cell to the mother so

Answer 20

a current has been generated and chemical energy has been converted into electrical energy

Question 21

half cells contain

Answer 21

chemical species present in a redox half equation

Question 22

example of how half cell is written for cu

Answer 22

cu2+(aq)|cu (s) half cellw

Question 23

what is the line in a half cell “ | “ referred to as

Answer 23

the phase boundary

Question 24

why are half cells joined together

Answer 24

for electrons to flow through wires from negative to positive

Question 25

zn has a - to - than cu

Answer 25

zu has a greater tendency to lose electrons than cu so has a more negative electrode potential

Question 26

zincs more negative electrode potential means

Answer 26

eqm lies to the left as zn (s) -> zn2+ (aq) + 2e- it is OXIDISED

Question 27

coppers more positive electrode potential means

Answer 27

eqm lies to the right as cu2+ (aq) + 2e- -> cu (s) it is REDUCED

Question 28

standard electrode potential

Answer 28

tendency to be reduced and gain electrons it is the emc( voltage) connected to a standard hydrogen half cell under standard conditions of 298k, 100kpa and 1moldm-3

Question 29

more negative standard electrode potential value

Answer 29

greater tendency to lose electrons and VICE VERSA

Question 30

ion/ion half cells

Answer 30

redox system does not contain a metal

Question 31

salt bridge

Answer 31

-usually made from a piece of filter paper soaked in a salt solution, usually potassium nitrate. - used to connect up the circuit -allows ions to move and conduct the charge