Assign oxidation numbers to all species in the redox reactions below. Identify the oxidizing and reducing agents.
1. Fe(OH)2 (s) + CrO42- (aq) → Fe2O3(s) + Cr(OH)4-(aq) (basic)
Fe is +2 on left and +3 on right. Cr is +6 on left and +3 on right
Oxidation is loss of electrons; the material oxidized is the reducing agent.
Reduction is gain of electrons; the material reduced is the oxidizing agent.
To assign oxidation numbers to the species in the redox reaction, we need to follow some rules:
1. The oxidation number of an atom in its elemental form is always zero. For example, Fe in Fe(OH)2(s) has an oxidation number of zero.
2. The sum of the oxidation numbers in a neutral compound is always zero. For example, in Fe(OH)2(s), the sum of the oxidation numbers of Fe and the two oxygen atoms (-2 each) should add up to zero.
3. The oxidation number of oxygen in most compounds is -2. However, in peroxides (such as H2O2) and when bonded with fluorine (such as OF2), the oxidation number of oxygen is -1.
4. The oxidation number of hydrogen is typically +1, except when it is bonded to metals, in which case it is -1.
Now let's assign the oxidation numbers to the species in the given redox reaction:
Fe(OH)2(s): Since oxygen has an oxidation number of -2 and we have two OH- groups, the oxidation number of Fe can be calculated as follows:
2(-2) + (-1) = -4 - 1 = -5
So, the oxidation number of Fe in Fe(OH)2(s) is +2.
CrO42-(aq): Since the overall charge of the CrO42- ion is -2, we can calculate the oxidation number of Cr as follows:
4(-2) + x = -2, where x is the oxidation number of Cr
-8 + x = -2
x = +6
Therefore, the oxidation number of Cr in CrO42-(aq) is +6.
Fe2O3(s): Since oxygen has an oxidation number of -2 and there are three oxygen atoms in Fe2O3, the oxidation number of Fe can be calculated as follows:
3(-2) + 2x = 0, where x is the oxidation number of Fe
-6 + 2x = 0
2x = 6
x = +3
So, the oxidation number of Fe in Fe2O3(s) is +3.
Cr(OH)4-(aq): Since the overall charge of the Cr(OH)4- ion is -1, the oxidation numbers can be calculated as follows:
4(-2) + x + 1(-1) = -1, where x is the oxidation number of Cr
-8 + x - 1 = -1
x = +6
Therefore, the oxidation number of Cr in Cr(OH)4-(aq) is +6.
Now, to identify the oxidizing and reducing agents, we need to compare the changes in oxidation numbers:
The oxidation number of Fe changes from +2 to +3, which means it has been oxidized (increased its oxidation number), so Fe is the reducing agent.
The oxidation number of Cr changes from +6 to +6, which means it has not been reduced or oxidized, so it is neither the reducing nor the oxidizing agent.
In summary, Fe is the reducing agent, and Cr is neither the reducing nor the oxidizing agent in the given redox reaction.