U13 A2 Learning aim B: Investigate oxidation-reduction reactions in order to understand their many applications in analysis

Oxidation/Reduction equation worksheet


For 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.

Redox Reaction Not a Redox Reaction
Oxidation
Reduction



1. NaCl (aq) + AgNO3 (aq)  AgCl (s) + NaNO3 (aq)

LHS: OS
Na=+1
Cl=-1
Ag=+1
N=+5
O=-2


RHS: OS
Na=+1
Cl=-1
Ag=+1
N=+5
O=-2


Not a Redox Reaction



2. H2 (aq) + F2 (aq)  2HF (aq)

LHS:
H2= 0
F2=0

RHS:
2H=+1
F=-1

H2 = 2H oxidation
F2=F reduction

Redox reaction

3. PCl5 (aq) + 4H2O (l)  H3PO4 (aq) + 5HCl (aq)


LHS:

1

, P= +5
Cl5=-1
4H2=+1
O=-2

P+ (5 x Cl) = 0
P+ (5x-1)=0
P+(-5)=0
P=+5

RHS:
H3=+1
P= +5
O4=-2
5H= +1
Cl=-1

(3xH) + P + (4 x 0) = 0
(3) + P (4x-2) = 0
(3) + P (-8) = 0
P + (-5) = 0
P = +5

Not a Redox Reaction


4. Zn (s) + 2HCl (aq)  ZnCl2 (aq) + 2H2 (g)


LHS:
Zn=0
2H=+1
Cl=-1

RHS:
Zn=+2
Cl2=-1
2H2=+0

Zn + (2xCl)=0
Zn+(2x-1)=0
Zn +(-2)=0
Zn=+2


Redox Reaction

5. Ca(OH)2 (aq) + 2HCl (aq)  CaCl2 (aq) + 2H2O (l)


LHS:
Ca =+2
O=-2
H=+1
2H=+1

2

, Cl=-1

RHS:
Ca=+2
Cl2=-1
2H2=+1
O=-2


Not a Redox Reaction


6. 3CuS (s) + 8HNO3 (aq)  3CuSO4 (aq) + 8NO (g) + 4H2O (l)




LHS:
3Cu=+2
S=-2
8H=+1
N=+5
O3=-2

H+N+(3x0)=0
H+N+(3x-2)=0
H+N+(-6)=0
N-5=0
N=+5

RHS:
3Cu=+2
S=+6
O=-2
8N=+2
O=-2
4H2=+1
O=-2

Redox Reaction



7. KMnO4 (aq) + 6H2SO4 (aq)  2K2SO4 (aq) + 4MnSO4 (aq) + 6H2O (l) + 5O2 (g)


LHS:
K=+1
Mn=+7
O4=-2
6H2=+1
S=+6
O4=-2



3

, RHS:
2K2=+1
S=+6
O4=-2
4Mn=+2
S=+6
O4=-2
6H2=+1
O=-2
5O2=0

Mn+S+(4x0)=0
Mn+(+6)+(4x-2)=0
Mn+(+6) + (-8)=0
Mn-2=0
Mn=+2


Redox Reaction

Balance the following redox equations, by identifying and using the oxidation numbers
within each species and the numbers of electrons lost or gained.


7. Fe2+  Fe3+ + e- (Oxidation - Oxidation number of Fe changed from +2 to
+3. Lost one electron)

MnO4- + 8H+ + 5e-  Mn2+ + 4H2O (Reduction - Oxidation number of Mn
changed from +7 to +2. Gained 5
electrons)

5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq)  5 Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)




8. IO3- + 6H+ + 6e-  I- + 3H2O (Reduction - Oxidation number of I
changed from +5 to -1. Gained 6
electrons)


2S2O32-  S4O62- + 2e- (Oxidation - Oxidation number of S changed from
+2 to +2.5. Lost two electrons)

IO3- (aq) + 6H+ (aq) + 6S2O32- (aq)  I- (aq) + 3S4O62- (aq) + 3H2O (l)




9. I2 + 2e-  2I- (Reduction - Oxidation number of I changed from 0 to -1.
Gained 2 electrons)


2S2O32-  S4O62- (aq) + 2e- (Oxidation - Oxidation number of S changed from
+2 to +2.5. Lost two electrons)


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