Download WEEK 6- Balancing Redox Reactions

ELECTROCHEMISTRY
(Refer to Chapter 19 – Electrochemistry)
OXIDATION – __________ of electrons by an atom, molecule, ion, etc.
NOTES:
REDUCTION – __________ of electrons by an atom, molecule, ion, etc.
e.g. 2Na(s) + Cl2(g) → 2NaCl(s)
Here the Na metal is _______________ to Na+ (1 electron lost per sodium atom)
and
Cl2 is _______________ to 2Cl– (1 electron gained per chlorine atom).
Aside: You may have heard the phrase: "LEO the lion says GER "
where LEO stands for "Loss of Electrons is Oxidation" and
GER stands for "Gain of Electrons is Reduction".
OR
OIL RIG – where "OIL" stands for "Oxidation Is Loss (of
electrons)" and "RIG" stands for "Reduction Is Gain (of electrons)"
Question: All the following are oxidation-reduction reactions EXCEPT
A.
H2(g) + F2(g) → 2HF(g).
B.
Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).
C.
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g).
D.
6Li(s) + N2(g) → 2Li3N(s).
E.
Ca(s) + H2(g) → CaH2(s).
Oxidation– Reduction or REDOX Reactions
Reactions which involve _________ GAIN (reduction) and LOSS (oxidation) of
electrons. It is convenient to think of the overall REDOX reaction as two individual
HALF-rxns, namely an oxidation half– reaction and a reduction half– reaction, i.e.,
+ Cl2(g)
2Na(s) → 2Na+(aq) + 2e–
+ 2e– → 2Cl– (aq)
(oxidation ½ – reaction)
(reduction ½– reaction)
2Na(s) + Cl2(g) → 2NaCl(aq)
Question: In the following, which species is oxidized?
3Ag2S(s) + 8H+(aq) + 2NO3–(aq) → 6Ag+(aq) + 3S(s) + 2NO(g) + 4H2O(l)
A.
B.
C.
D.
E.
NO3–
Ag2S
NO
H+
Ag+
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Oxidising AGENT – causes another species to be _______________ by
NOTES:
accepting/taking electrons from it.
Reducing AGENT – causes another species to be _______________
reduced by donating electrons to it.
“LEORA GEROA”
LEORA → Loss of Electrons is Oxidation and that species is a Reducing Agent.
GEROA → Gain of Electrons is Reduction and that species is an Oxidising Agent.
Oxidation results in ____________ in the Oxidation State (or Number) by loss of
electrons. Reduction results in _____________ in the Oxidation State (or Number)
by gain of e–’s.
Question: In the following redox reaction,
8H+(aq) + 6Cl–(aq) + Sn(s) + 4NO3–(aq) → SnCl62–(aq) + 4NO2(g) + 4H2O(l),
the oxidizing agent is
A.
Cl–.
B.
H+.
C.
Sn.
D.
NO3–.
E.
SnCl62–.
Rules for assigning Oxidation Numbers
(Refer to Section 4.5 to review oxidation and reduction reactions.)
1. For pure elements, their oxidation number = 0 (e.g., Fe(s), Br2(l) and N2(g), etc.).
2. Monatomic ions have oxidation numbers equal to their charge (e.g., Fe3+ or S2–).
3. Fluorine in its compounds is ALWAYS –1
4. Hydrogen in covalent compounds with non– metals is +1 (e.g., H2O, CH4, etc.),
in metal hydrides it is –1 (e.g., NaH, CaH2, etc.)
5. Oxygen in its covalent compounds is –2, except in peroxides (e.g., H2O2) where
it is –1 (rule #4) or in metal oxides such as KO2 where O is –½.
6. The sum of the oxidation numbers equal zero for a neutral (no charge)
compound; for an ion the sum of the oxidation numbers equals the ion’s
charge.
7. Look to the column number in the periodic table to assign oxidation numbers.
Note: As shown in rule #5, non-integral oxidation numbers are possible.
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Examples: Assign Oxidation Numbers (or States) to all the atoms in the following:
NOTES:
NH3
CO2
Cr2O72–
C2H6O
LiH
C2H4O2
S4O62–
I3 –
Other Examples (answers at end of notes):
PbS
NO3–
AlO2–
CH4
H3PO4
ClO–
SF6
H2PO4–
ClO2–
MnO4–
F2O
ClO4–
Absolutely FOOL– PROOF method for balancing ANY redox equation
(This method is different from the textbook’s method.)
1.
Assign Oxidation Numbers and write separate equations for each half
reaction, then work on each half– reaction separately.
2.
Balance the atoms being oxidised or reduced.
3.
Based on the change in oxidation numbers, determine the number of e–’s
involved and add them to the high oxidation number side of the equation.
4.
Balance the charge, adding either H+ (if in ACID solution) or OH– (if in
BASE solution).
5.
Balance the OXYGENs by adding H2O to the appropriate side of the eq’n.
6.
Check that everything balances – it will!
7.
Scale the two half rxns so the number of e–’s are equal in each equation.
8.
Add the two half– reactions.
To remember this order, use "AEIOU":
A – after assigning oxidation numbers, check that the Atoms are balanced.
E – add the Electrons to the half reactions.
I – balance the Ions (charge) using either H+ or OH– .
O – balance the Oxygens in the half reactions by adding H2O.
U – Unite the two equations together after scaling the two half reactions.
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Example 1 : MnO4– (aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq) in acid solution.
NOTES:
Oxidation numbers: Mn in MnO4– is ___, in Mn2+ it is ____; Fe2+ is +2, Fe3+ is +3
Oxidation (Loss of e-):
Fe2+(aq) → Fe3+(aq) + 1e– (steps 1,2,3)
Reduction (Gain of e-):
MnO4– (aq) → Mn2+(aq)
(steps 1 and 2)
(step 3)
(step 4)
(step 5)
Visual check – all OK (step 6)
Scale (step 7):
8H+ + 5e–
5Fe2+(aq) → 5Fe3+(aq) + 5e–
+ MnO4– (aq) → Mn2+(aq) + 4H2O(l)
(step 7)
(step 7)
Add the two half reactions together and cancel out the electrons (step 8):
8H+(aq) + MnO4– (aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Example 2 : In base, NO2– (aq) + Al(s) → NH3(g) + AlO2–(aq)
Oxidation Numbers:
N in NO2–(aq) is ______, in NH3(g) is _______
Al in Al(s) is ______, in AlO2– (aq) is _____
Oxidation:
Al(s) → AlO2–(aq) + 3e–
(steps 1,2,3)
(steps 4)
(steps 5,6)
Reduction:
NO2–(aq) → NH3(g)
(steps 1 and 2)
(step 3)
(step 4)
(steps 5,6)
Scale (step 7):
8OH–(aq) + 2Al(s) → 2AlO2–(aq) + 4H2O(l) + 6e–
NO2– (aq) + 6e– + 5H2O(l) → NH3(g) + 7OH–(aq)
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Add:
5H2O(l) + NO2–(aq) + 8OH– (aq) + 2Al(s) → NH3(g) + 7OH–(aq) + 2AlO2–(aq) + 4H2O(l)
Tidy up (cancel superfluous water and hydroxide ions):
H2O(l) + NO2–(aq) + OH–(aq) + 2Al(s) → NH3(g) + 2AlO2–(aq)
Practice: Balance the following reactions.
1. H+(aq) + Cr2O72–(aq) + C2H6O(l) → Cr3+(aq) + CO2(g)
2. As2O3(s) + NO3–(aq) → H3AsO4(aq) + NO(g)
3. MnO4
–
(aq)
+ S
2–
(aq)
→ MnS(s) + S(s)
4. Cl2(g) → Cl–(aq) + ClO–(aq)
6. Ag2S(s) + CN
(aq)
7. V(s) → HV6O17
–
8. CNS
(aq)
(in acid)
(in base)
(in base)
5. HO2– (aq) + Cr(OH)3–(aq) → CrO42– (aq)
–
(in base)
+ O2(g) → S(s) + [Ag(CN)2]– (aq)
3–
(aq)
(H+ in equation signifies acidic)
+ H2(g)
(in acid)
(in base)
+ O2(g) → CO2(g) + NO2(g) + SO2(g)
(in acid)
Qu#4: All the following half-reactions are balanced EXCEPT
A.
2H2O + 2e– → H2 + 2OH–.
B.
2Ta + 5H2O → Ta2O5+ 10H+ + 10e–.
C.
NO3 + 4H++ + 3e– → NO + 2H2O.
D.
H2O2 → 2OH– + 2e–.
E.
H3PO3 + H2O → H3PO4 + 2H+ + 2e–.
Answers to assigning oxidation number examples on page 68:
PbS
S (like O) is –2; ∴Pb is +2 since compound is neutral.
CH4
H = +1; ∴C = –4 since compound is neutral.
SF6
F = –1; ∴S = +6 since compound is neutral.
MnO4–
NO3
–
O = –2, ∴ to solve for N: x + 3(–2) = –1; x = +5
H3PO4
H2PO4
O = –2, ∴ to solve for Mn use: x + 4(–2) = –1; x = +7
–
O = –2, H = +1; ∴ to solve for P: 3 + x + 4(–2) = 0; x = +5
O = –2, H = +1; ∴ to solve for P: 2 + x + 4(–2) = –1; x = +5
F = –1, ∴ O = +2 (Rule 3) since compound is neutral.
F2O
AlO2–
O is –2, ∴Al = +3 since ion must add up to –1.
–
O is –2, ∴ Cl = +1 since ion must add up to –1.
ClO2
–
O is –2, ∴ Cl = +3 since ion must add up to –1.
ClO4
–
O is –2, ∴ Cl = +7 since ion must add up to –1.
ClO
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Answers for Balancing Practice (page 70):
1. 16H+(aq) + 2Cr2O72–(aq) + C2H6O(l) → 4Cr3+(aq) + 2CO2(g) + 11H2O(l)
2. 3As2O3(s) + 4NO3–(aq) + 4H+(aq) + 7H2O(l) → 6H3AsO4(aq) + 4NO(g)
3. 2MnO4– (aq) + 7S2–(aq) + 8H2O(l) → 2MnS(s) + 5S(s) + 16OH–(aq)
4. Cl2(g + 2OH–(aq) → Cl–(aq) + ClO–(aq) + H2O(l)
Note: In this example, Cl2 is used as a reactant in BOTH the oxidisation and reduction rxns, i.e.,
Oxidation: Cl2(g) → 2ClO–(aq);
Reduction: Cl2(g) → 2Cl–(aq)
Also, the equation is divided by 2 to create the simplest expression.
5. 2HO2– (aq) + Cr(OH)3–(aq) → CrO42– (aq) + OH–(aq) + 2H2O(l)
(Since a basic solution, use either OH– or H2O as the product to balance the reduction half reaction.)
6. 2Ag2S(s) + 8CN–(aq) + O2(aq) + 4H+(aq) → 2S(s) + 4[Ag(CN)2]– (aq) + 2H2O(l)
(Since under acidic conditions, one uses H2O as the product to balance the reduction half reaction.
Also keep "weird" atoms together, i.e., the CN– must be in the half reaction with [Ag(CN)2]– .)
7. 6V(s) + 14H2O(l) + 3OH– (aq) → HV6O173 –(aq) + 15H2(g)
8. CNS– (aq) + O2(g) → CO2(g) + NO2(g) + SO2(g)
(in acid)
To determine oxidation numbers for CNS– (aq) use your knowledge of electronegativity:
Assign S (same group as O) as –2 and therefore C+N = +1. To get a sum of +1, C = +4 & N = –3
works (These assignments are based on which element is more electronegative. But, one can also
assign C = –4 & N = +5 and end up with the correct balanced equation at the end.)
Also, in this example, more than one element is being oxidized (i.e., both N & S). Just put them all in
the same half reaction and sum up the total number of electrons lost, to get the complete half reaction,
i.e.,
Oxidation half reaction: CNS– (aq) → CO2(g) + NO2(g) + SO2(g)
Ox. numbers: +4,– 3,– 2
+4
+4
+4
CNS– (aq) → CO2(g) + NO2(g) + SO2(g) + (7+6)e–
CNS– (aq) + 6H2O(l) → CO2(g) + NO2(g) + SO2(g) + 12H+(aq) + 13e–
Reduction half reaction: O2(g) + 4H+(aq) + 4e– → 2H2O(l)
Balanced overall equation:
4CNS– (aq) + 13O2(g) + 4H+(aq) → 4CO2(g) + 4NO2(g) + 4SO2(g) + 2H2O(l)
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