Electrochemistry Made Easy –
Essential Study Guide & Problem-
Solving Strategies
In which of these compounds does phosphorous have an oxidation number of +5?
A. PH3
B. CoPO4
C. Mn3(PO4)2
D. PCl3
E. P2O5 - B, C, E
Why B is Correct-
First, remember that polyatomic ions in the compound must have oxidation numbers
that equal the overall charge. PO4 has a 3- negative charge so the oxidation
numbers of phosphorus and oxygen must equal negative 3. Using known rules,
Oxygen has an oxidation number of 2-. Multiply -2 x 4 because there are 4 oxygen's
in the ion to get a total charge of -8. Since P + O4 must equal -3, phosphorus must
have an oxidation number of =+5.
Co= +3
P= +5
O= -2
Why C is correct-
It is important to recognize the subscript changes the amount of elements in the
compound. Using the known rules, we know that O is -2 and Mn, since it is in group
2 would be +2. Multiplying through with the subscript. we get:
Mn= 2 x 3= +6 and
O= -2 x 4 x 2= -16
To calculate the oxidation number of Phosphorus, remember that there are 2 so that
should be accounted for.
+6 + 2x + -16= 0
x must be equal to +5
Why E is correct-
O = -2 x 5= -10
P= 2x= 10 to equal 0. X= -5
The oxidation number of Mn in Mn3(PO4)2 is - +2
, It is important to recognize the subscript changes the amount of elements in the
compound. Using the known rules, we know that O is -2 and Mn, since it is in group
2 would be +2.
Mn= 2 x 3= +6 and
O= -2 x 4 x 2= -16
P= 2 x 5= 10
The oxidation number of hydrogen in AlH3 is - -1
AlH3
H= -1 x 3= -3
Al= +3
In Sn (s) + 2HCl (aq) ⟶ SnCl2 (aq) + H2 (g), what is being oxidized and what is
being reduced? - Sn is being oxidized, H2 is being reduced.
1. Assign oxidation numbers to all atoms in the equation and ignore the coefficients
in the equation.
Sn (s) + 2HCl (aq) ⟶ SnCl2 (aq) + H2 (g)
0 +1 -1 +2 -1 0
2. Compare oxidation numbers from the reactant side to the product side of the
equation. If a redox reaction has occurred, you will find that the oxidation numbers of
two (no more/no less) elements have changed from the reactant side to the product
side. The element oxidized is the one whose oxidation number increased. The
element reduced is the one whose oxidation number decreased.
Sn= 0 ⟶ Sn = +2, tin is oxidized
H= +1 ⟶ H= 0=, hydrogen is reduced
Is ZnCO3 (s) ⟶ ZnO (s) + CO2 (g) a redox reacrtion? - No, nothing is being oxidized
or reduced.
2 Ga (l) + 3 Br2 (l) ⟶ 2 GaBr3 (s)
In this reaction, what is the oxidizing agent and what is the reducing agent? - Br2 is
the oxidizing agent
Ga is the reducing agent
1. Assign oxidation numbers to each compound in the reaction.
2 Ga (l) + 3 Br2 (l) ⟶ 2 GaBr3 (s)
0 0 +3 -1
2. Compare oxidation numbers from reactant to products.
Ga= 0 ⟶ Ga= +3, Gallium is oxidized, therefore it is the reducing agent since the
electrons it loses reduced the charge of Bromine.
Br= 0 ⟶ Br= -1, Bromine is reduced, therefore it is the oxidizing agent since the
electrons it gains are a result of Gallium losing electrons.
For this reaction, which is the correct unbalanced oxidation half reaction?
Essential Study Guide & Problem-
Solving Strategies
In which of these compounds does phosphorous have an oxidation number of +5?
A. PH3
B. CoPO4
C. Mn3(PO4)2
D. PCl3
E. P2O5 - B, C, E
Why B is Correct-
First, remember that polyatomic ions in the compound must have oxidation numbers
that equal the overall charge. PO4 has a 3- negative charge so the oxidation
numbers of phosphorus and oxygen must equal negative 3. Using known rules,
Oxygen has an oxidation number of 2-. Multiply -2 x 4 because there are 4 oxygen's
in the ion to get a total charge of -8. Since P + O4 must equal -3, phosphorus must
have an oxidation number of =+5.
Co= +3
P= +5
O= -2
Why C is correct-
It is important to recognize the subscript changes the amount of elements in the
compound. Using the known rules, we know that O is -2 and Mn, since it is in group
2 would be +2. Multiplying through with the subscript. we get:
Mn= 2 x 3= +6 and
O= -2 x 4 x 2= -16
To calculate the oxidation number of Phosphorus, remember that there are 2 so that
should be accounted for.
+6 + 2x + -16= 0
x must be equal to +5
Why E is correct-
O = -2 x 5= -10
P= 2x= 10 to equal 0. X= -5
The oxidation number of Mn in Mn3(PO4)2 is - +2
, It is important to recognize the subscript changes the amount of elements in the
compound. Using the known rules, we know that O is -2 and Mn, since it is in group
2 would be +2.
Mn= 2 x 3= +6 and
O= -2 x 4 x 2= -16
P= 2 x 5= 10
The oxidation number of hydrogen in AlH3 is - -1
AlH3
H= -1 x 3= -3
Al= +3
In Sn (s) + 2HCl (aq) ⟶ SnCl2 (aq) + H2 (g), what is being oxidized and what is
being reduced? - Sn is being oxidized, H2 is being reduced.
1. Assign oxidation numbers to all atoms in the equation and ignore the coefficients
in the equation.
Sn (s) + 2HCl (aq) ⟶ SnCl2 (aq) + H2 (g)
0 +1 -1 +2 -1 0
2. Compare oxidation numbers from the reactant side to the product side of the
equation. If a redox reaction has occurred, you will find that the oxidation numbers of
two (no more/no less) elements have changed from the reactant side to the product
side. The element oxidized is the one whose oxidation number increased. The
element reduced is the one whose oxidation number decreased.
Sn= 0 ⟶ Sn = +2, tin is oxidized
H= +1 ⟶ H= 0=, hydrogen is reduced
Is ZnCO3 (s) ⟶ ZnO (s) + CO2 (g) a redox reacrtion? - No, nothing is being oxidized
or reduced.
2 Ga (l) + 3 Br2 (l) ⟶ 2 GaBr3 (s)
In this reaction, what is the oxidizing agent and what is the reducing agent? - Br2 is
the oxidizing agent
Ga is the reducing agent
1. Assign oxidation numbers to each compound in the reaction.
2 Ga (l) + 3 Br2 (l) ⟶ 2 GaBr3 (s)
0 0 +3 -1
2. Compare oxidation numbers from reactant to products.
Ga= 0 ⟶ Ga= +3, Gallium is oxidized, therefore it is the reducing agent since the
electrons it loses reduced the charge of Bromine.
Br= 0 ⟶ Br= -1, Bromine is reduced, therefore it is the oxidizing agent since the
electrons it gains are a result of Gallium losing electrons.
For this reaction, which is the correct unbalanced oxidation half reaction?
