Electrochemistry Test Questions with Correct Answers Graded A+
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 - Answers 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 - Answers +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 - Answers -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? - Answers
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? - Answers 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? - Answers 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?
, Al (s) + Ni^2+ (aq) ⟶Ni (s) + Al^3+ (aq)
A. Al (s) ⟶ Al^3+ (aq)
B. Ni^2+ (aq) ⟶ Al^3+ (aq)
C. Ni (s) ⟶ Ni^2+ (aq)
D. Ni^2+ ⟶ (aq) Ni (s)
E. Al (s) ⟶ Ni (s)
F. Al^3+ (aq) ⟶ Al (s) - Answers A. Al (s) ⟶ Al3+ (aq)
1. Assign oxidation numbers to all the elements in the reaction.
Al (s) + Ni2+ (aq) ⟶ Ni (s) + Al3+ (aq)
0 2+ 0 3+
2. Aluminum is being oxidized so that will be the oxidation half reaction.
Al (s) ⟶ Al^3+ (aq)
For this reaction, what is the coefficient of Al (s) in the overall balanced equation?
Al (s) + Ni^2+ (aq) ⟶Ni (s) + Al^3+ (aq)
A. 1
B. 2
C. 3
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 - Answers 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 - Answers +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 - Answers -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? - Answers
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? - Answers 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? - Answers 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?
, Al (s) + Ni^2+ (aq) ⟶Ni (s) + Al^3+ (aq)
A. Al (s) ⟶ Al^3+ (aq)
B. Ni^2+ (aq) ⟶ Al^3+ (aq)
C. Ni (s) ⟶ Ni^2+ (aq)
D. Ni^2+ ⟶ (aq) Ni (s)
E. Al (s) ⟶ Ni (s)
F. Al^3+ (aq) ⟶ Al (s) - Answers A. Al (s) ⟶ Al3+ (aq)
1. Assign oxidation numbers to all the elements in the reaction.
Al (s) + Ni2+ (aq) ⟶ Ni (s) + Al3+ (aq)
0 2+ 0 3+
2. Aluminum is being oxidized so that will be the oxidation half reaction.
Al (s) ⟶ Al^3+ (aq)
For this reaction, what is the coefficient of Al (s) in the overall balanced equation?
Al (s) + Ni^2+ (aq) ⟶Ni (s) + Al^3+ (aq)
A. 1
B. 2
C. 3
