Electrochemistry Study Preparation with Complete Solutions: Exercises and Explanations, Exams of Electrochemistry

Electrochemistry Study Preparation with Complete
Solutions: Exercises and Explanations, Exams of
Electrochemistry


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 - ANSWER-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 - ANSWER-+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 - ANSWER--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? -
ANSWER-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? - ANSWER-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? - ANSWER-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) - ANSWER-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.