TEACHERS’ COPY ON 2015 TUTORIAL FOR REDOX PART I : OXIDATION NUMBERS AND REDOX EQUATIONS Level 1 1. Write down the oxidation oxidation numbers of : (a) Chlorine in Cl2, POCl3, HClO, NaClO2, ClO2, ClO3-, and Cl2O7 Cl2
POCl3
HClO
NaClO2
ClO2
ClO3-
Cl2O7
0
-1
+1
+3
+4
+5
+7
(b) Nitrogen in N2, N2O, NO, NO2, NO3-, N2H4 and HCN -
N2
N2O
NO
NO2
NO3
N2H4
HCN
0
+1
+2
+4
+5
-2
-3
2. Write redox equations for the following reactions. (a)
When sodium sulfite, Na2SO3 is added to acidified potassium dichromate( VI), K2Cr2O7 solution, the orange solution turns to green Cr3+ ions. Oxidation ½ eqn: SO32- + H2O SO42- + 2H+ + 2eReduction ½ eq: Cr2O72- + 14H+ + 6e2Cr3+ + 7H2O Eqn(1) x 3 + Eqn(2) gives
(b)
3SO32- + Cr2O72- + 8H+ orange
3SO42- + 2Cr3+ + 4H2O green
When aqueous potassium iodide, KI is added to acidified hydrogen peroxide, H 2O2, a brown solution of iodine is observed. Oxidation ½ eqn: 2II2 + 2e Reduction ½ eqn: H2O2 + 2H+ + 2e-
--- (1) --- (2)
2H2O +
Eqn(1) + Eqn(2) gives 2I + H2O2 + 2H
(c)
--- (1) --- (2)
+ 2H2O (brown solution of iodine) I2
Sodium nitrite, NaNO2 decolourises acidified potassium manganate( m anganate(VII), KMnO4. Oxidation ½ eqn: NO2- + H2O NO3- + 2H+ + 2e--- (1) + 2+ Reduction ½ eqn: MnO4 + 8H + 5e Mn + 4H2O --- (2) -
-
+
Eqn(1) x 5 + Eqn(2) x 2 gives 5NO2 + 2MnO4 + 6H
purple (d)
5NO3- + 2Mn2+ + 3H2O pale pink
Chlorine gas, Cl2, reacts with cold sodium hydroxide solution to form sodium chloride and sodium chlorate(I), NaClO. Oxidation ½ eqn: Cl 2 + 4OH2Cl O- + 2H2O + 2e--- (1) Reduction ½ eqn: Cl 2 + 2e 2Cl --- (2) Cl O + Cl + H2O Eqn(1) + Eqn(2) gives Cl 2 + 2OH 1
Level 2 3. The equations for three reactions are given below: Cl2 + 2H2O + SO2 Cl2 + H2S SO2 + 2H2S
2HCl + H2SO4 2HCl + S 2H2O + 3S
What is the correct order of strength of the three reacting gases as reducing agents? Weakest
Strongest
A
chlorine
sulfur dioxide
hydrogen sulfide
B
chlorine
hydrogen sulfide
sulfur dioxide
C
sulfur dioxide
hydrogen sulfide
chlorine
D
hydrogen sulfide
sulfur dioxide
chlorine
Ans: A O.A. R.A. Cl2 + 2H2O + SO2 O.A. R.A. Cl2 + H2S O.A. R.A. SO2 + 2H2S
2HCl + H2SO4
2HCl + S 2H2O + 3S
H2S is the strongest reducing agent since it reduces both SO 2 and Cl 2. Cl 2 is the weakest reducing agent since it cannot reduce both SO 2 and H2S. SO2 is the intermediate in strength as a reducing agent since it reduces Cl 2 but not H2S. 4. When iron is reacted with aqueous iron(III) ions, iron(II) ions are formed. Assuming the reaction goes to completion, how many moles of Fe and Fe 3+(aq) would result in a mixture containing equal numbers of moles of Fe3+(aq) and Fe2+(aq) once the reaction had taken place? [N07/ I /2]
A B C D
moles of Fe 1 1 1 2
moles of Fe +(aq) 2 3 5 3
Ans: C Oxidation ½ eqn: Fe Fe2+ + 2eReduction ½ eqn: Fe3+ + eFe2+
--- (1) --- (2)
Eqn(1) + Eqn(2) x 2 gives a balanced redox eqn of Fe + 2Fe3+
3Fe2+
Hence, 1 mol Fe will react with 2 mol Fe3+ to produce 3 mol Fe 2+. --THE END-2
TEACHERS’ COPY ON 2015 TUTORIAL FOR REDOX PART II: REDOX REACTIONS IN VOLUMETRIC ANALYSIS Level 2 Potassium Manganate(VII) & Potassium Dichromate(VI) Titration 1. Compounds of iron and strontium are usually found in food. Spinach is considered to be a rich source of iron. However, it contains iron absorption inhibiting substances such as oxalate which can bind to the iron to form iron( II) oxalate, which renders much of the iron in spinach unusable by the body. 35.0 cm3 sample of iron(II) oxalate, FeC2O4 was extracted from 150 g of spinach, diluted in water and the solution made up to 250 cm 3. A 25.0 cm3 portion of this solution was acidified and required 13.45 cm3 of 0.020 mol dm-3 potassium manganate(VII) for oxidation of iron(II) to iron(III) and oxalate ions to carbon dioxide. Solution: (a) State the change in oxidation number for manganese and carbon in the reaction. Mn: C:
(b)
From +7 to +2 [or decrease by 5] From +3 to +4 [or increase by 1]
Write down all the relevant ionic half equations and hence the overall redox equation for the reaction between potassium manganate(VII) and iron(II) oxalate. MnO4- + 8H+ + 5eMn2+ + 4H2O C2O422CO2 + 2eFe2+ Fe3+ + e3MnO4- + 24H+ + 5C2O42- + 5Fe2+
(c)
3Mn2+ + 5Fe3+ + 12H2O + 10CO2
Calculate the concentration, in mol dm-3, of iron(II) oxalate in the original sample. 13.45 x 0.020 1000 = 2.69 x 10-4 mol 5 Number of moles of FeC2O4 in 25.0 cm3 = x 2.69 x 10-4 3 = 4.48 x 10-4 mol 250 Number of moles of FeC2O4 in 250 cm3 = 4.48 x 10-4 x 25.0 -3 = 4.48 x 10 mol 4.48 x 10 -3 Concentration of FeC2O4 in the original sample = 35.0 x 10 -3 = 0.128 mol dm-3 Number of moles of KMnO4
=
3
Iodine – Thiosulfate Titration 2. The halogens form many interhalogen compounds and ions, in which a halogen atom lower down the group is surrounded by atoms of halogens higher in the group. Two such species are the compound C lF5 and the ion ICl4-. Solution: (a) Calculate the oxidation number of iodine in ICl4-.
[1]
+3 (b)
N-chlorosuccinimide (M r = 133.5) is used in some water purification tablets. It dissolves in water to give a weak acid, HC lO, which kills waterborne bacteria.
N-chlorosuccinimide The amount of N-chlorosuccinimide in a tablet can be estimated by dissolving it in water, and adding an excess of acidified K I(aq). The HClO is quantitatively reduced to Cl- by I- ions, which are oxidised to I2. The I2 is titrated with standard sodium thiosulfate. The following reaction takes place during the titration. 2-
I2 + 2S2O3
(i)
→
2I- + S4O62-
Construct a balanced equation for the reaction between HClO and acidified KI(aq). HCl O + 2I- + H+
→
I2 +
Cl - + H2O
When one tablet was subjected to the above procedure, 6.0 cm 3 of 0.0050 mol dm-3 sodium thiosulfate was required to discharge the colour of I2. (ii)
Calculate the mass of N-chlorosuccinimide in the tablet. n S O 2 2
−
=
0.0050 ×
3
6.0 1000
=
[4]
3.00 × 10 -5 mol
1 -5 -5 × 3.00 × 10 = 1.50 × 10 mol 2 n N-chlorosuccinimide = n HClO = n I 2 = 1.50 × 10 -5 mol n I2
=
m N-chlorosuccinimide
=
1.50 × 10 -5 × 133.5 = 2.00 × 10
−
3
g [N11/ III /1]
4
Further Redox Reactions 3. The compound Na3CrO4, is an instable dark green solid. The addition of dilute sulfuric acid to this produces a solution containing chromium(III) ions and dichromate(VI) ions. When a 1.00 g sample of the impure green solid was reacted in this way, the Cr(VI) in the resulting solution reacted with an excess of potassium iodide to liberate 5.00 x 10-3 moles of iodine, I2.
Solution: (a) Calculate the oxidation state of chromium in Na3CrO4, and construct a balanced equation for its reaction with dilute acid. O.N of Cr in Na3CrO4
= +5
3CrO43- + 10H+ Cr3+ + Cr2O72- + 5H2O 6Na3CrO4 + 10H2SO4 Cr2(SO4)3 + 2Na2Cr2O7 + 10H2O + 7Na2SO4 (b)
(ionic eqn) (full eqn)
Write an equation for the reaction between Cr(VI) and iodide ions, and hence calculate the number of moles of Cr(VI) produced in solution.
Cr2O72- + 14H+ + 6INumber of moles of
3I2 + 2Cr3+ + 7H2O
I2 produced
= 5.00 x 10 -3 mol
1 x 5.00 x 10-3 3 = 1.67 x 10-3 mol
Number of moles of Cr2O72- =
number of moles of Cr(VI) produced
(c)
= 2 x 1.67 x 10-3 = 3.34 x 10-3 mol
By using your results in (a) and (b), calculate the number of moles of Na3CrO4, in the 1.00 g sample and hence the percentage purity.
Number of moles of Na3CrO4
=
6 moles of Cr2O722
Mass of Na3CrO4 in 1 g sample = 3 x 1.67 x 10-3 x 185.0 = 0.925 g 0.925 x 100% 1 = 92.5 %
% purity of Na3CrO4 sample =
[N96/ II /2]
5
Level 3 4. 1.50 g of a copper(II) compound was dissolved in water and then made up to 250 cm3. 25.0 cm3 portion of this solution were treated with excess potassium iodide solution. Iodine and solid copper(I) iodide were formed. The iodine liberated was titrated with a standard sodium thiosulfate solution containing 2.80 g of thiosulfate ions per dm3. The average titre was 24.00 cm3. Calculate the percentage of copper(II) in the compound. [Titration video: http://vimeo.com/18901164. Password: vjcchemistry]
Solution: 2S2O32- + I2 S4O62- + 2INumber of moles of S2O32- used
Number of moles of
I2
evolved
= conc. (mol dm-3) x volume = [2.8 (2(32.0) + 3(16.0))] x 0.024 = 6.00 x 10-4 mol = ½ x 6.00 x 10-4 = 3.00 x 10-4 mol
2Cu2+ + 4I2CuI + I2 Number of moles of Cu2+ in 25.0 cm3 = 2 x 3.00 x 10-4 = 6.00 x 10-4 mol Number of moles of Cu2+ in 250 cm3
250 25 .0 -3 = 6.00 x 10 mol
= 6.00 x 10-4 x
Mass of Cu(II) in 1.5 g of compound = 6.00 x 10-3 x 63.5 = 0.381 g Percentage of Cu(II) in compound
0.381 x 100% 1.5 = 25.4 %
=
5. In one experiment, 30.0 cm3 of a solution of M5+ of concentration 0.100 mol dm-3 was reduced by sulfur dioxide to a solution of Mn+. In order to reoxidise Mn+ to M5+, 50.0 cm3 of acidified potassium dichromate(VI) solution of concentration 0.0200 mol dm-3 was required. What is the value of n?
Solution: [O] half-eqn: Mn+ M5+ + (5 - n)e[R] half-eqn: Cr2O72- + 14H+ + 6e2Cr3+ + 7H2O 50.0 -3 = 1.00 × 10 mol n Cr O 2 = 0.020 × 2 7 1000 n e = 6 × (1.00 × 10 3 ) = 6.00 × 10 -3 mol −
-
n M5
+
=
0.10 ×
30.0 1000
=
3.00 × 10 -3 mol
6
n e
n M3
= +
6.00 × 10 3.00 × 10
3 =
3
2
1 mol M5+ will take in 2 mol e -. i.e. M5+ + 2e n = 3
M3+
6. Iodine and chlorine react together to form compound X (ICln). When 0.0010 mol of X was reacted with an excess of K I (aq), all of its iodine was converted into I2. The iodine liberated required 40.0 cm 3 of 0.10 mol dm-3 sodium thiosulfate, Na2S2O3 for complete reaction. Solution: (a) Calculate the amount (in moles) of iodine produced. I2
+ 2S2O32-
2I- + S4O62-
Amount of iodine produced
(b)
Hence calculate the value of n in ICln. ICl n
+
KI
n
→
0.001 mol
n I 2 n ICl n
=
n =
(c)
1 40.0 × × 0.10 2 1000 = 0.00200 mol
=
0.002 0.001
[(1 + n)/2] I2 + nKCl 0.002 mol
=
(1 + n ) / 2 1
3
Write a balanced equation for the reaction between Cl2 and I2.
Substitute n = 3 into the equation ‘ I2 + nCl2 2ICl 3 I2 + 3Cl 2
2ICl n’
→
[N03/ III /5 either]
--THE END—
7
ANSWERS TO 2015 TUTORIAL SUPPLEMENTARY QUESTIONS FOR OXIDATION NUMBERS AND REDOX EQUATIONS 3. Write down the oxidation numbers of : Manganese in Mn2(CO)10, MnCl2, MnF3, MnO2, Na3MnO4, KMnO4 Mn2(CO)10
MnCl2
MnF3
MnO2
Na3MnO4
KMnO4
0
+2
+3
+4
+5
+7
2. In each of the following reactions, state what (if any) has been reduced or oxidised. Write ion-electron half equations (where appropriate) to explain your answer. (a) 3Cl2 + 6OH – → 5Cl – + ClO3 – + 3H2O Cl 2 is reduced and oxidised.
Cl 2 + 12OH2Cl O3- + 6H2O + 10eCl 2 + 2e2Cl -
(b)
(ON from 0 to +5) (ON from 0 to -1)
HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4 – No redox
3. Write redox equations for the following reactions. (a)
Chlorine dioxide, ClO2 disproportionates to a mixture of chlorate(III), ClO2- and chlorate(V), ClO3- in alkaline solution.
Oxidation ½ eqn: Reduction ½ eqn:
Cl O2 + 2OHCl O3- + H2O + eCl O2 + eCl O2-
Eqn(1) + Eqn(2) gives
(b)
2Cl O2 + 2OH-
Cl O2- + Cl O3- + H2O
NH2OH is oxidised to N2O by Fe3+(aq), which is itself reduced to Fe2+(aq).
N2O + H2O + 4H+ + 4e- Fe2+
Oxidation ½ eqn: 2NH2OH Reduction ½ eq: Fe3+ + eEqn(1) + Eqn(2) x 4 gives
(c)
----(1) ----(2)
4Fe3+ + 2NH2OH
----(1) ----(2)
4Fe2+ + N2O + H2O + 4H+
Reaction of potassium iodide, KI with aqueous potassium manganate (VII), KMnO4 gives a mixture of black and brown precipitate. The black precipitate dissolves in excess KI solution. Oxidation ½ eqn: 2II2 + 2e Reduction ½ eqn: MnO4- + 2H2O + 3e-
MnO2 + 4OH -
8
----(1) ----(2)
3I2 + 2MnO2 + 8OHblack brown
Eqn(1) x 3 + Eqn(2) x 2 gives 6I + 2MnO4 + 4H2O
Black ppt of I2 dissolves in excess KI solution: I2 +
-
I
-
I3
-
5. The head of the first self-igniting match, called a ‘lucifer’, contained a mixture of antimony(III) sulfide and potassium chlorate(V). When the match was ‘struck’, the following reaction took place Sb2S3(s) + 3KClO3(s) → Sb2O3(s) + 3KCl(s) + 3SO2(g) Which element is reduced in the reaction?
A B C D
antimony chlorine oxygen sulfur [H1 N10/ I /10]
Ans: B ON of chlorine decreases from +5 in KCl O3 to -1 in KCl 6. Ten percent of the copper produced in the USA comes from bacterial leaching of lowgrade copper ores. In this process acidified water is sprayed onto the ore chalcopyrite. Bacteria then convert the insoluble ore into a solution containing iron and copper ions. 4CuFeS2 + 17O2 + 4H+ → 4Cu2+ + 4Fe3+ + 8SO42- + 2H2O No change occurs in the oxidation state of copper. What changes in oxidation state occur for the iron and the sulfur in this reaction? change in oxidation state Fe
S
A
-1
-6
B
-1
+6
C
+1
-8
D
+1
+8 [H1 N09/ I /9]
Ans: D ON of iron increases from +2 in CuFeS2 to +3 in Fe3+; ON of sulfur increases from -2 in CuFeS2 to +6 in SO42-
9
7. When aqueous hydrogen peroxide, H2O2, is mixed with acidified potassium dichromate(VI), there is a colour change from orange to green. When aqueous hydrogen peroxide is added to acidified potassium iodide solution, there is a colour change from colourless to brown. The oxidation number of oxygen in H2O2 is -1. What are the oxidation numbers of oxygen after the reactions with potassium dichromate(VI) and potassium iodide? [N09/ I /16]
A B C D
After reaction with potassium dichromate(VI) -2 -2 0 0
After reaction with potassium iodide -2 0 -2 0
Ans: C K2Cr2O7 oxidises H2O2 to O2 (ON: 0) while KI reduces H2O2 to H2O (ON: -2).
10
ANSWERS TO 2015 TUTORIAL SUPPLEMENTARY QUESTIONS FOR REDOX REACTIONS IN VOLUMETRIC ANALYSIS 1. Sodium sulfite acts as a preservative in sausages. The amount of sulfite present in a sausage can be determined by boiling the sausage with a strong acid, and measuring the amount of sulfur dioxide evolved. 50 g of sausage together with 200 cm3 of water and 20 cm3 of 10 mol dm-3 HCl are boiled in a distillation flask. The sulfur dioxide gas liberated is dissolved in water in collecting flask. The resulting solution in the collecting flask required 7.00 cm3 of 0.0250 mol dm-3 iodine solution in order to completely oxidise the sulfur dioxide gas to sulfate(VI) ion. As a check on the titration, excess BaCl2 solution is then added and the resulting precipitate collected and weighed. (a) Give an ion-electron half-equation for the oxidation reaction of sulfur dioxide to sulfate(VI) ion and hence write the balanced chemical equation for the reaction between sulfur dioxide and iodine solution. [2]
SO2 + 2H2O SO42- + 4H+ + 2e2II2 + 2e Overall: SO2 + 2H2O + I2 SO42- + 4H+ + 2I(b) Calculate the mass of the sulfur dioxide gas evolved from 50 g of sausage.
[2]
Amount of SO2 = Amount of I2 7.00 = 0.0250 x 1000 -4 = 1.75 x 10 mol Mass of SO2 = 1.75 x 10-4 x [32.1 + 2(16.0)] = 0.0112 g (c) Express the mass of sulfur dioxide in sausage parts per million (ppm).
[1]
0.0112 x106 50 = 224 ppm
Mass of SO2 in sausage =
(d) Identify and calculate the mass of the precipitate formed when BaCl2 solution is added to the solution at the end point of the titration? [2]
Precipitate is BaSO4. Amount of BaSO4
= Amount of SO42- = Amount of SO2 = 1.75 x 10-4 mol
Mass of BaSO4 = 1.75 x 10-4 x [137.3 + 32.1 + 4(16.0)] = 0.0408 g
11
2.
Wines often contain a small amount of sulfur dioxide that is added as a preservative. The sulfur dioxide content of a wine is found using the following method: A 50 cm3 sample of white wine is reacted with 40.0 cm3 of 0.0100 mol dm-3 of excess aqueous iodine. The sulfur dioxide in the wine is oxidized to sulfate(VI) ion in the process. The unreacted iodine requires exactly 23.60 cm3 of 0.0200 mol dm-3 of sodium thiosulfate for complete reaction.
(a) Give the ion-electron half-equation for the oxidation reaction of sulfur dioxide to sulfate(VI). Hence, write the redox equation for the reaction between sulfur dioxide and aqueous iodine. [2]
SO2 + 2H2O SO42- + 4H+ + 2e2II2 + 2e Overall: SO2 + 2H2O + I2 SO42- + 4H+ + 2I(b) Write the redox equation for the reaction between aqueous iodine and sodium thiosulfate. [1]
2S2O32- +
I2
S4O62- + 2I -
(c) Determine the concentration of sulfur dioxide, in mol dm-3, in the 50 cm3 sample of wine. [3]
Amount of unreacted
1 x amount of S 2O322 1 23 .60 = × 0.0200 × 2 1000 -4 = 2.36 x 10 mol
I2 =
40.0 1000 = 4.00 x 10-4 mol
Amount of original I2 = 0.0100 ×
Amount of reacted
4.00 x 10 -4 – 2.36 x 10-4 = 1.64 x 10-4 mol
I2 =
amount of SO2 = amount of reacted
I2 =
1.64 x 10 -4 mol
1.64 × 10 4 × 1000 Concentration of SO2 in sample = 50 = 3.28 x 10-3 mol dm-3 −
3. Chlorine dioxide, ClO2 oxidises thiosulfate, S2O32- under acidic conditions. ClO2(aq) + 4H+(aq) + 5e-
→
Cl-(aq) + 2H2O(l)
In an experiment, 25.0 cm3 portion of the solution containing 11.2 g of thiosulfate ions per dm3 were oxidised by 20.00 cm3 of 0.0250 mol dm-3 of chlorine dioxide in acidic medium. 12
(a) State the oxidation state of sulfur in thiosulfate ion.
[1]
+2 (b) Calculate the new oxidation number of sulfur after the reaction.
[3]
Conc. of S2O32-
= 11.2 / [2(32.1) + 3(16.0)] = 0.0998 mol dm-3 25.0 Amount of S2O32- = 0.0998 × 1000 = 2.50 x 10-3 mol 20 .00 1000 -4 = 5.00 x 10 mol = 5 x 5.00 x 10-4 = 2.50 x 10-3 mol
Amount of Cl O2 = 0.0250 × Amount of e-
n ∴
e
_
nS O 2
=
2.50 × 10 2.50 × 10
−
2 3
_
3
_
3
=
1
Therefore, 1 mol S2O32- will give out 1 mol ei.e. 1 mol of S in S2O32- will give out 0.5 mol eHence, new oxidation number of sulfur = +2.5 4. An acidified solution of the salt KClOx (x = 1,2 or 3) will oxidise Fe 2+(aq) to Fe3+(aq) quantitatively, the chlorine being reduced to Cl-(aq). When 0.150 g of the salt KClOx was reacted with 0.500 mol dm-3 Fe2+ (aq) in the presence of H+ (aq), 11.30 cm3 of Fe2+ (aq) was needed for complete reaction. Calculate the value of x and construct an equation for the reaction between Fe 2+(aq) and acidified KClOx (aq). [N11/ III /5f] Oxidation ½ eqn: Fe2+ Fe3+ + e- Reduction ½ eqn: Cl Ox- + 2x H+ + 2x e→
n Fe
=
2+
n e
n ClO
−
0.500 ×
-
=
x
n -
=
11 .30 1000
=
Cl + xH2O
→
5.65 × 10 - 3 mol
1× (5.65 × 10 -3 ) = 5.65 × 10 -3 mol
0.150 mol 39.1 + 35 .5 + 16 x 5.65 × 10 3 = 2 x 0.150 39.1 + 35.5 + 16 x −
e
n ClO
= −
x
-
= 2
x
13
----(1) ----(2)
Oxidation ½ eqn: Fe2+ Fe3+ + e- ----(1) + Reduction ½ eqn: Cl O2 + 4H + 4e Cl + 2H2O ----(2) 2+ + 3+ 4Fe + Cl - + 2H2O Eqn(1) x 4 + Eqn(2) gives 4Fe + Cl O2 + 4H →
→
→
--THE END--
14
