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Summary Unit 13 - Applications of Inorganic Chemistry BTEC Unit 13 Assignment 2 all criteria
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Unit 13 - Applications of Inorganic Chemistry BTEC Unit 13 Assignment 2 all criteria / distinction received
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Oxidation/Reduction equation worksheetFor the following six reactions, identify the oxidation numbers of the elements in the elements and
compounds involved.
From this, determine which equations are examples of redox reactions and which are not.
1. NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
LHS :OS RHS: OS
Na= +1 Na = +1
Cl= -1 Cl = -1
Ag= +1 Ag =+1
N=+5 N =+5
O=-2 O =-2
No change = not a redox reaction .
2. H2 (aq) + F2 (aq) → 2HF (aq)
LHS :OS RHS: OS
H= 0 H = +1
F= 0 F = -1
This is a redox reaction:
H = oxidizing ( oxidisation occurring)
F = reducing (reduction occurring)
3. PCl5 (aq) + 4H2O (l) → H3PO4 (aq) + 5HCl (aq)
LHS :OS RHS: OS
P = +5 P = +5
Cl= -1 Cl= -1
H = +1 H = +1
O = -2 O = -2
No change = this is not a redox reaction.
4. Zn (s) + 2HCl (aq) → ZnCl2 (aq) + 2H2 (g)
LHS :OS RHS: OS
Zn = 0 Zn = +2
H = +1 H =0
Cl = -1 Cl = -1
This is a redox reaction:
Zn = oxidizing ( oxidisation occurring )
H = reduction ( reduction occurring )
5. Ca(OH)2 (aq) + 2HCl (aq) → CaCl2 (aq) + 2H2O (l)
LHS :OS RHS: OS
Ca = +2 Ca = +2
O = -2 O = -2
H = +1 H = +1
Cl = -1 Cl = -1
No change = this is not a redox reaction.
6. 3CuS (s) + 8HNO3 (aq) → 3CuSO4 (aq) + 8NO (g) + 4H2O (l)
LHS :OS RHS: OS
Cu = +2 Cu = +2
S = -2 S = +6
H = +1 H = +1
BTEC Applied Science Level 3 / Unit 13 Assignment 2 / Answered by Moreen Mero
,N = +5 N = +2
O = -2 O = -2
This is a redox reaction:
S = oxidizing ( oxidisation occurring )
N = reducing (reduction occurring )
7. KMnO4 (aq) + 6H2SO4 (aq) → 2K2SO4 (aq) + 4MnSO4 (aq) + 6H2O (l) + 5O2 (g)
LHS :OS RHS: OS
K = +1 K = +1
Mn = +7 Mn = +2
O = -2 O = -2
H = +1 H = +1
S = +6 S = +6
This is a redox reaction:
Mn = reducing ( reduction occurring )
O = oxidizing ( oxidisation occurring )
Balance the following redox equations, by identifying and using the oxidation numbers within each species
and the numbers of electrons lost or gained.
8. …Fe2+ (aq) + …MnO4- (aq) + …H+ (aq) → …Fe3+ (aq) + …Mn2+ (aq) + …H2O (l)
LHS :OS RHS: OS
Fe = +2 Fe = +3
Mn = ? Mn = +2
Finding the oxidisation of Mn =
Mn + (O x 4 ) = -1
Mn + (-2 x 4) = -1
Mn + ( -8 ) = -1
Mn = +7
Oxidisation half equation :
( Fe2+ → Fe3+ + e- ) x 5
Reduction half equation :
8H+ + MnO4- + 5e- → Mn2+ + 2H2O
Redox full equation :
5Fe2+ + 8H+ + MnO4- + 5e → 5Fe3+ + 5e- + Mn2+ + 4H2O
Final equation:
5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O
9. …IO3- (aq) + …H+ (aq) + …2S2O32- (aq) → …I- (aq) + …S4O62- (aq) + …H2O (l)
LHS :OS RHS: OS
I =? I = -1
S =? S =?
O = -2 O = -2
Finding oxidation of I (IO3-) :
I + (Ox3) = -1
BTEC Applied Science Level 3 / Unit 13 Assignment 2 / Answered by Moreen Mero
,I + (-2x3) = -1
I + (-6) = -1
I = +5
Finding S2O3 :
(Sx2) + (Ox3) = -2
(Sx2) + (-2x3) = -2
(Sx2) + (-6) = -2
(Sx2) = 2+6
Sx2 = +4
S =+2
Finding reduction of S (S4O62- ) :
(Sx4) + (Ox6) = -2
(Sx4) + (-2x6) = -2
(Sx4) + (-12) = -2
Sx4 = +10
S = 2.5
Oxidation half equation:
( 3S2O3-2 → S4O6-2 + 2e- ) x3
Reduction half equation:
6H+ + IO3- + 6e- → I- + 3H2O
Redox full equation:
6S2O3-2 + 6H+ + IO3- → 3S4O6-2 + I- + 3H2O
10. …S2O32- (aq) + …I2 (aq) → …I- (aq) + …S4O62- (aq)
LHS :OS RHS: OS
S = +2 S = 2.5
O = -2 O = -2
I 0 I = -1
Oxidation half equation :
2S2O3-2 → S4O6-2 + 2e-
Reduction half equation :
I2 + 2e- → 2I-
Redox half equation :
2S2O3-2 + I2 → S4O6-2 + 2I-
BTEC Applied Science Level 3 / Unit 13 Assignment 2 / Answered by Moreen Mero
, Half Equations and REDOX Equations
Write out the half equations and redox equations for the following cells:
o Zn(s) | Zn2+(aq) ¦¦ Cu2+(aq) | Cu(s)
Oxidation half equation: Zn(s) → Zn2+(aq) + 2e-
Reduction half equation: Cu2+(aq) + 2e- → Cu(s)
Full redox equation: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Cell Voltage Calculation:
Eocell = Eoreduction + E°oxidation
= 0.76 + 0.34 = 1.10V
o Zn(s) | Zn2+(aq) ¦¦ Pb2+(aq) | Pb(s)
Oxidation half equation: Zn(s) → Zn2+(aq) + 2e-
Reduction half equation: Pb2+(aq) + 2e- → Pb(s)
Full redox equation: Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb (s)
Cell Voltage Calculation:
Eocell = Eoreduction + E°oxidation
= 0.76 – 0.13 = 0.43 V
o Pb(s) | Pb2+(aq) ¦¦ Cu2+(aq) | Cu(s)
Oxidation half equation: Pb(s) → Pb2+(aq) + 2e-
Reduction half equation: Cu2+(aq) + 2e- → Cu(s)
Full redox equation: Pb(s) + Cu2+(aq) → Pb2+(aq) + Cu (s)
Cell Voltage Calculation:
Eocell = Eoreduction + E°oxidation
= 0.13 + 0.34 = 0.47 V
BTEC Applied Science Level 3 / Unit 13 Assignment 2 / Answered by Moreen Mero
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