chem keynote 2

8/11/2019 Chem Keynote 2

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REDOX

In a redox reaction, reduction and oxidation occur together:

• the oxidation number of one element goes up - oxidation • the oxidation number of another element must go down - reduction • the total increase in oxidation number of the element oxidised must balance

the total decrease in oxidation number of the element reduced.

Mg (s) + 2HCl (aq) ———> MgCl! (aq) + H! (g)

Mg 0 ———-> +2

H (2x) +1 ———-> 0



REDOX TITRATIONS 1

Find the balanced equation for the oxidation of

an iron (II) salt by potassium manganate (VII)

in acid solution.

Half equations:

Fe!" (aq) ——> Fe"" (aq) + e#

MnO$# (aq) + 8H" (aq) +5e# —-> Mn!" (aq) + 4H!O (l)

• The reaction above can be used in redox titrations to estimate the amount

of a reducing agent (iron (II) in iron tablets).• KMnO$# is a powerful oxidising agent and has intense purple colour in (aq).• In acid, it is reduced to almost colour manganese (II) salts.• This titration is self-indicating.• The end-point is when all iron (II) has been oxidised to iron (III). Manganate

(VII) gives a permanent pink colour.• Reaction stoichiometry gives mole ratio of 5mol Fe!" to 1 mol MnO$#.



REDOX TITRATIONS 2

2S!O%!# (aq) + I! (aq) —-> 2I# (aq) + S$O&!# (aq)

•

Iodine in potassium iodide is dark brown.• Addition of Na!S!O% results in light brown/yellow colour

as iodine is reduced giving a colourless solution at the

end point.• Near the end-point, while solution is pale yellow,

starch indicator is added giving blue-black solution.

The end-point change is now blue to colourless.• Used to find concentration of oxidising agent that

oxidises iodide ions to iodine.

To estimate the % of Cu in alloy, the Cu alloy is dissolved, and reacted with

excess KI to produce Iodine. Liberated Iodine is titrated with Na!S!O%.

• Weighed alloy reacted with nitric acid, giving solution containing Cu!" (aq).• 2Cu!" (aq) + 4I#(aq) ——> 2CuI (s) + I! (aq).• Liberated iodine is estimated by titrating with Na!S!O%.• Overall stoichiometry: 2mol Cu!"= 1mol I! , reduced by 2mol S!O%!#.• Hence, 1:1 mole ratio between Cu!" and S!O%!#.



Quick Questions!

1. In the analysis of the percentage of copper in an alloy, explain why an excess

of potassium iodide must be used.

2. What is the oxidation number of chromium in a chromate ion, CrO$!# ?

3. Hydrogen iodide, HI, is oxidised to iodine, I!, by concentrated sulfuric acid,

H!SO$, which is itself reduced to hydrogen sulfide, H!S. Write a balanced full

equation for this reaction.



1. To ensure that all the copper (II) reacts, and that the number of moles of

thiosulfate in the titration with iodine corresponds to the number of moles ofcopper.

2. +6

3. 8HI (g) + H !SO $ (l) ——-> H !S (g) + 4I ! (s) + 4H !O (l)

Quick Answers!



Hydrogen and Alcohol Fuel Cells

Half equations:

2H!O (l) + 2e# ——-> H! (g) + 2OH# (aq) E cell = -0.83V

0.5O! (g) + H!O (l) + 2e# ——> 2OH# (aq) E cell = +0.4V

(-)

(+)



Fuel-Cell Breathalysers

f

3C!H'OH (g) + 2Cr!O"#$ (aq) +16H" (aq) ——> 3CH%COOH + 4Cr%& (aq) +14H!O (aq)

• Sensitivity was limited

• Amount of ethanol in breath depends onconcentration in blood, “mouth alcohol”

can give errors in readings.• Drivers brought to station for further test• IR analysis, ethanol (2950cm##, C-H)• OH group is not used because water in

breath affects reading.

• Platinum electrode oxidises ethanol to

ethanoic acid, releasing p and e# •

p’s combine with oxygen on other side,forming water.• Alcohol (, Oxidation (, Current (

chem keynote 2

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