(BLUE) REDOX II Flashcards
(BLUE) REDOX II
infocard read and recite
Redox is concerned with the movement of electrons, we can divide this movement up into two processes. An oxidation reaction which is a loss of electrons and a reduction reaction which is a gain of electrons . These reactions can never appear on their own, as both a source and sink for the electrons is always required. This is why they are called redox reactions We combine the two equations to make a redox equation. These reactions can be followed by looking at the change in oxidation number. If the oxidation number increases the species has undergone oxidation. If the oxidation number decreases the species has undergone reduction. The oxidation number itself is defined as the charge the species would have if the bonding was 100% ionic.
In some reactions we use reagents which are particularly good at gaining or losing electrons e.g. KMnO4 in acid is a good oxidising agent. It is able to oxidise other species, by accepting electrons; the Mn(VII) ion itself is reduced to Mn(II). A good reducing agent is Zn, it is very good at donating electrons; the Zn itself is oxidised to Zn2+
info card read and recite
How to write and use half equations.
· Identify atoms that change using oxidation numbers, write out their oxidation numbers above each atom.
· Balance with electrons – Remember OIL RIG
· Balance with H+ / H2O
· Check charges are equal either side. This gives the balanced half equations for oxidation and reduction
· Multiply up one or both ½ equations so that the electrons cancel
· Add equations, cancelling electrons and any H+ / H2O This results in a Total redox equation
what does oil rig stand for
oil :
loss of e -
increase in oxidation number
gain of oxygen , loss if hydrogen
rig
gain of e-
decrease in oxidation number
loss of oxygen / gain of hydrogen
5) Which of the following are redox reactions?
a) Mg + 2 HCl —> MgCl2 + H2
b) MgO + 2 HCl —-> MgCl2 + H2O
c) Al2O3 + 2 Fe —> 2 Al + Fe2O3
d) [Co(H2O)6]2+ + 4 Cl- —> [CoCl4]2- + 6 H2O
A) redox
B) not redox
C) redox
D) not redox
Write out the equation for the oxidation of Fe2+ to Fe3+.
Fe2+ —-> Fe3+ + e-
Write out the equation for the reduction of MnO4- to Mn2+ (remember to add acid)
MnO4- + 8H+ + 5e- —–> Mn2+ + 4H2O
Write out to the full redox equation of MnO4- to Mn2 and Fe2+ to Fe3+.
5Fe2+ + MnO4- + 8H+ —-> 5Fe3+ + Mn2+ + 4H2O
1) In a titration 25cm3 of iron(II) sulfate required 22.4 cm3 of 0.02 mol dm-3 potassium manganate(VII) solution at the end point. What is the concentration of the solution of iron(II) sulfate?
This equation shows that for every mole of manganate (VII) you need …5.. moles of Fe2+
Calculate the moles of potassium manganate
Deduce the moles of iron(II) required for reaction and hence the concentration of iron(II) sulfate
Moles = 22.40/1000 x 0.02 = 4.48 x 10^-4
Moles Fe = 5 x 4.48 x 10-4 = 2.24 x 10^-3
Concentration = moles x 1000/volume = 2.24 x 10^-3 x 1000/25 = 0.0896 mol dm-1
2) 25cm3 of a solution of 0.02 mol dm-3 of ethanedioic acid, H2C2O4 , reacted with 20cm3 of 0.01 mol dm-3 of potassium manganate(VII). How many moles of ethanedioic acid react with 1 mole of
potassium manganate(VII). Suggest a likely equation for the reaction.
Moles MnO4- = 20/1000 x 0.01 = 0.20 x 10^-3
Moles ethandioate = 25/1000 x 0.02 = 0.50 x 10-
Molar ratio MnO4- : ethandioate = 2:5
therefore
MnO4- + 8H+ + 5e- —-> Mn2+ + 4H2O (x2)
H2C2O4 —–> 2CO2 + 2H+ 2e- ( x5)
2MnO4- 6H+ + 5H2C2O4 —-> 2Mn2+ + 8H2O + 10CO2
info card read and recite
Potassium manganate(VII) is a well-known oxidising agent, usually used in solutions acidified with dilute sulfuric acid. Solutions containing MnO4-(aq) ions have an intense purple colour, whereas those containing Mn2+(aq) ions are almost colourless. Solutions containing Fe2+(aq) ions can be titrated against potassium manganate(VII) solution. The colour of the manganate(VII) is discharged, the end-point of the titration being the point at which the addition of one more drop of potassium manganate(VII) gives a permanent pale purple colour.
This titration forms the basis of an analytical technique for the estimation of iron in the double salt you made in the L6.
info card read and recite
There are a large number of reactants that can oxidize iodide to iodine, therefore if you add an excess of iodide to a known quantity of an oxidizing agent, iodine is produced. The amount of iodine can be measured by titrating with sodium thiosulfate solution. If the amount of iodine produced is known then you can then work out the number of moles of the original oxidizing agent.
what is the Reaction of iodine with thiosulfate
2S2O32- + I2 —-> S4O62- + 2I-
A household bleach contains sodium chlorate(I), NaOCl, the chlorate(I) ion will react with iodide to give iodine:-
ClO- + 2I- + 2H+ —> I2 + Cl- + H2O
A 25cm3 sample of household bleach was diluted to 250 cm3. A 25cm3 portion of this was added to an excess of potassium iodide and titrated against 0.2 M sodium thiosulphate. The volume required was 18.5 cm3. What is the concentration of sodium chlorate(I) in the bleach?
a) Calculate the moles of thiosulfate used in the titration
b) Deduce the number of moles of iodine liberated from original reaction with bleach
c) Deduce the number of moles of chlorate ions in the 25cm3 portion
d) Calculate the concentration of the diluted bleach in the volumetric\flask
f) Calculate the concentration of the concentrated bleach
a) moles thio = 18.5/1000 x 0.20 = 3.7 x 10^-3
b) Molar ratio thio : iodine = 2 : 1
Moles I2 =( 3.7 x 10^-3)/2 = 1.85 x 10-3
c) Molar ratio iodine : chlorate = 1 : 1
Moles OCl- = 1.85 x 10^-3
d) Conc dil OCl- = mole x 1000/vol = 1.85 x 10^-3 x 1000/25 = 0.074 mol dm-3
f) Concentrated bleach = 0.074 x 250/25(sample) = 0.74 mol dm-3
info card read and recite
Recent hospital outbreaks of MRSA and Norovirus indicate the need for thorough disinfection. Usually bleach is used on surfaces – current guidelines indicate 0.5% (w/v) hypochlorite for routine use.
Household bleach is about 5% (w/v) sodium hypochlorite. Diluting neat bleach from the supermarkets by 1 part in 10 should give a solution of this concentration of 0.5%.
Bleach is made by dissolving chlorine in sodium hydroxide solution:-
C12(aq) + 2OH-(aq) <—-> OCl-(aq) + Cl-(aq) + H2O(1)
You will carry out an analysis of supermarket bleach to find out if the diluted solution gives 0.5% (w/v) sodium hypochlorite (NaOCl).
An aqueous solution of bleach, sodium chlorate(I), is fully ionised in solution:
NaOCl(aq) —–> Na+(aq) + OCl-(aq)
In acidic conditions, these chlorate(I) ions act as an oxidising agent, or electron acceptor. They will oxidise iodide ions to iodine:-
Reduction half equation: OCl-(aq) + 2H+(aq) + 2e- —-> Cl-(aq) + H2O(1)
Oxidation half equation: 2I- —> I2 + 2e-
Total redox reaction: ClO- + 2I- + 2H+—–> I2 + Cl- + H2O
The iodine can be estimated by titrating it with sodium thiosulphate solution:
Oxidation half equation: 2S2O32- —> S4O62- + + 2e-
Reduction half equation: I2 + 2e- —-> 2I-
Total redox reaction: 2S2O32- + I2 —> S4O62- + 2I-
The iodine colour disappears at the endpoint. This can be made sharper by the addition of starch.
