For each of the following reactions, state whether it is an oxidation or a reduction. Then, for each one, write the balanced half-reaction.
a. Fe³+ (aq) → Fe²+ (aq)
b. 2O²- (aq) → O₂ (g)
c. Al³+ (aq) → Al(s)
d. Fe²+ (aq) → Fe(s)
e. Mg (s) → Mg²+ (aq)
Straight forward. Just remember the definitions. Oxidation is the loss of electrons. Reduction is the gain of electrons. Here is a memory aid. LEO the lion goes GER. LEO = loss electrons oxidation. GER = gain electrons reduction.
a. Fe³+ (aq) → Fe²+ (aq)
Fe^3+ + e ==> Fe^2+ that's a GER so reduction.
b. 2O²- (aq) → O₂ (g)
2O2^2- ==> 2O2 +2e that's a LEO so oxidation
I shall be happy to check the others if you post your answers.
a. The reaction is a reduction because Fe³+ is gaining electrons to become Fe²+.
The balanced half-reaction is:
Fe³+ (aq) + e⁻ → Fe²+ (aq)
b. The reaction is an oxidation because O²- is losing electrons to form neutral O₂.
The balanced half-reaction is:
2O²- (aq) → O₂ (g) + 4e⁻
c. The reaction is a reduction because Al³+ is gaining electrons to form solid Al.
The balanced half-reaction is:
Al³+ (aq) + 3e⁻ → Al (s)
d. The reaction is a reduction because Fe²+ is gaining electrons to form solid Fe.
The balanced half-reaction is:
Fe²+ (aq) + 2e⁻ → Fe (s)
e. The reaction is an oxidation because Mg is losing electrons to form Mg²+.
The balanced half-reaction is:
Mg (s) → Mg²+ (aq) + 2e⁻
To determine whether a reaction is an oxidation or a reduction, we need to identify the changes in the oxidation states of the elements involved.
An increase in the oxidation state indicates oxidation, while a decrease indicates reduction.
Let's analyze each reaction:
a. Fe³+ (aq) → Fe²+ (aq)
In this reaction, Fe³+ is being reduced to Fe²+. The oxidation state of Fe is decreasing from +3 to +2. Therefore, the reaction is a reduction.
To write the balanced half-reaction, we need to balance the number of atoms and the charges on each side:
Half-reaction: Fe³+ (aq) + e⁻ → Fe²+ (aq)
b. 2O²- (aq) → O₂ (g)
In this reaction, O²- is being oxidized to O₂. The oxidation state of O is increasing from -2 to 0. Therefore, the reaction is an oxidation.
To write the balanced half-reaction, we need to balance the number of atoms and the charges on each side:
Half-reaction: 4O²- (aq) → 2O₂ (g) + 4e⁻
c. Al³+ (aq) → Al(s)
In this reaction, Al³+ is being reduced to Al. The oxidation state of Al is decreasing from +3 to 0. Therefore, the reaction is a reduction.
To write the balanced half-reaction, we need to balance the number of atoms and the charges on each side:
Half-reaction: Al³+ (aq) + 3e⁻ → Al(s)
d. Fe²+ (aq) → Fe(s)
In this reaction, Fe²+ is being reduced to Fe. The oxidation state of Fe is decreasing from +2 to 0. Therefore, the reaction is a reduction.
To write the balanced half-reaction, we need to balance the number of atoms and the charges on each side:
Half-reaction: Fe²+ (aq) + 2e⁻ → Fe(s)
e. Mg (s) → Mg²+ (aq)
In this reaction, Mg is being oxidized to Mg²+. The oxidation state of Mg is increasing from 0 to +2. Therefore, the reaction is an oxidation.
To write the balanced half-reaction, we need to balance the number of atoms and the charges on each side:
Half-reaction: Mg(s) → Mg²+ (aq) + 2e⁻
Hope this helps! Let me know if you have any further questions.