2. Identify the oxidation numbers for each element in the following equations. Then determine whether each equation describes a redox reaction. Justify your answer in terms of electron transfer. Describe the oxidation and reduction that occurs.
a. Ca + 2HCl → CaCl2 + H2 (4 points)
b. 2KOH + K2Cr2O7 → 2K2CrO4 + H2O (4 points)
c. AgNO3 + KI → AgI + KNO3 (4 points)
d. 2PbS + 3O2 → 2PbO + 2SO2 (5 points)
a. In the equation Ca + 2HCl → CaCl2 + H2, the oxidation number of Ca is 0, the oxidation number of H in HCl is +1, and the oxidation number of Cl is -1. In CaCl2, the oxidation number of Ca is +2 and the oxidation number of Cl is -1. In H2, the oxidation number of H is 0.
This equation describes a redox reaction because there is a transfer of electrons. Ca is oxidized from an oxidation number of 0 to +2, while H in HCl is reduced from an oxidation number of +1 to 0.
b. In the equation 2KOH + K2Cr2O7 → 2K2CrO4 + H2O, the oxidation number of K is +1, the oxidation number of O in KOH is -2, the oxidation number of Cr in Cr2O7 is +6, and the oxidation number of O in Cr2O7 is -2. In K2CrO4, the oxidation number of K is +1, the oxidation number of Cr is +6, and the oxidation number of O is -2. In H2O, the oxidation number of H is +1 and the oxidation number of O is -2.
This equation describes a redox reaction because there is a transfer of electrons. The Cr in Cr2O7 is reduced from an oxidation number of +6 to +6, while the O in KOH is oxidized from an oxidation number of -2 to -2.
c. In the equation AgNO3 + KI → AgI + KNO3, the oxidation number of Ag is +1, the oxidation number of O in NO3 is -2, the oxidation number of N in NO3 is +5, the oxidation number of K is +1, the oxidation number of I is -1, and the oxidation number of N in KNO3 is +5. In AgI, the oxidation number of Ag is +1 and the oxidation number of I is -1.
This equation does not describe a redox reaction because there is no change in oxidation numbers for the elements involved. The Ag in AgNO3 and AgI remains at an oxidation number of +1, and the I in KI and AgI remains at an oxidation number of -1.
d. In the equation 2PbS + 3O2 → 2PbO + 2SO2, the oxidation number of Pb in PbS is +2 and the oxidation number of S in PbS is -2. In PbO, the oxidation number of Pb is +2 and the oxidation number of O is -2. In SO2, the oxidation number of S is +4 and the oxidation number of O is -2.
This equation describes a redox reaction because there is a transfer of electrons. The Pb in PbS is oxidized from an oxidation number of +2 to +2, while the S in PbS is reduced from an oxidation number of -2 to +4.
a. To identify the oxidation numbers for each element in the equation, we can assign them based on their usual oxidation states.
In Ca + 2HCl → CaCl2 + H2, Ca is in its usual +2 oxidation state. H always has an oxidation state of +1 in compounds, and Cl usually has an oxidation state of -1 in compounds.
In CaCl2, Ca has an oxidation state of +2, and Cl has an oxidation state of -1.
In H2, H has an oxidation state of 0 since it is in its elemental form.
Based on the assignment of oxidation numbers, we can see that Ca is oxidized from an oxidation state of 0 to +2. H is reduced from an oxidation state of +1 to 0. Therefore, this equation describes a redox reaction as there is a transfer of electrons.
The oxidation reaction is: Ca (0) → CaCl2 (+2)
The reduction reaction is: 2HCl (+1) → H2 (0)
b. In 2KOH + K2Cr2O7 → 2K2CrO4 + H2O, K has an oxidation state of +1, O typically has an oxidation state of -2, and H has an oxidation state of +1 in compounds.
In K2CrO4, K has an oxidation state of +1, Cr has an oxidation state of +6, and O has an oxidation state of -2.
In H2O, H has an oxidation state of +1, and O has an oxidation state of -2.
Based on the assignment of oxidation numbers, we can see that Cr is reduced from an oxidation state of +6 to +4. Therefore, this equation describes a redox reaction as there is a transfer of electrons.
The oxidation reaction is: K2Cr2O7 (+6) → 2K2CrO4 (+4)
The reduction reaction is: 2KOH (+1) → H2O (0)
c. In AgNO3 + KI → AgI + KNO3, Ag typically has an oxidation state of +1, N has an oxidation state of -3 in nitrate (NO3-) ions, and O typically has an oxidation state of -2.
In AgI, Ag has an oxidation state of +1, and I has an oxidation state of -1.
In KNO3, K has an oxidation state of +1, N has an oxidation state of +5, and O has an oxidation state of -2.
Based on the assignment of oxidation numbers, we can see that Ag is reduced from an oxidation state of +1 to 0. I is oxidized from an oxidation state of -1 to 0. Therefore, this equation describes a redox reaction as there is a transfer of electrons.
The oxidation reaction is: AgNO3 (+1) → AgI (0)
The reduction reaction is: KI (-1) → KNO3 (+5)
d. In 2PbS + 3O2 → 2PbO + 2SO2, Pb typically has an oxidation state of +2, S typically has an oxidation state of -2, and O typically has an oxidation state of -2.
In PbO, Pb has an oxidation state of +2, and O has an oxidation state of -2.
In SO2, S has an oxidation state of +4, and O has an oxidation state of -2.
Based on the assignment of oxidation numbers, we can see that S is oxidized from an oxidation state of -2 to +4. Therefore, this equation describes a redox reaction as there is a transfer of electrons.
The oxidation reaction is: 2PbS (-2) → 2SO2 (+4)
The reduction reaction is: 3O2 (0) → 2PbO (+2)
To identify the oxidation numbers for each element in the given equations, we need to know the rules for assigning oxidation numbers. Here are the guidelines:
1. The oxidation number of an element in its elemental form is zero. For example, Ca and O in CaO have oxidation numbers of 0.
2. The sum of the oxidation numbers in a neutral compound is zero.
3. The oxidation number of a monatomic ion is equal to its charge.
4. In most compounds, oxygen has an oxidation number of -2, unless it is in a peroxide where its oxidation number is -1.
5. In most compounds, hydrogen has an oxidation number of +1, unless it is in a hydride where its oxidation number is -1.
6. The sum of the oxidation numbers in a polyatomic ion is equal to its charge.
Let's go through each equation to determine the oxidation numbers:
a. Ca + 2HCl → CaCl2 + H2
- The oxidation number of Ca is +2 since HCl is a compound and the sum of the oxidation numbers is 0.
- The oxidation number of H in HCl is +1 since Cl has an oxidation number of -1.
- In CaCl2, Ca still has an oxidation number of +2, and Cl has an oxidation number of -1.
- In H2, H still has an oxidation number of +1.
Now, let's determine whether this equation describes a redox reaction. A redox reaction involves the transfer of electrons from one element to another. To do this, we can compare the oxidation numbers of the elements before and after the reaction:
- In Ca, the oxidation number is +2 both before and after the reaction.
- In H, the oxidation number is +1 before the reaction and 0 after the reaction.
Since there is a change in oxidation number for H, we can conclude that this equation describes a redox reaction. The H atoms are oxidized from +1 to 0, and Ca remains unchanged.
b. 2KOH + K2Cr2O7 → 2K2CrO4 + H2O
- The oxidation number of K is +1 since KOH is a compound and the sum of the oxidation numbers is 0.
- The oxidation number of O in KOH is -2.
- The oxidation number of Cr in K2Cr2O7 can be found by using the fact that the sum of oxidation numbers is zero in a neutral compound. Given that K has an oxidation number of +1 and O has an oxidation number of -2, we can calculate it as follows:
2x + 2(-2) + 7x = 0
2x - 4 + 7x = 0
9x - 4 = 0
9x = 4
x ≈ 0.44
The oxidation number of Cr is approximately +0.44 in K2Cr2O7.
- In K2CrO4, K still has an oxidation number of +1, and Cr has an oxidation number of +6.
- In H2O, H has an oxidation number of +1, and O has an oxidation number of -2.
Let's check whether this equation describes a redox reaction:
- In K, the oxidation number is +1 both before and after the reaction.
- In Cr, the oxidation number is +0.44 before the reaction and +6 after the reaction.
- In H, the oxidation number is +1 before the reaction and 0 after the reaction.
- In O, the oxidation number is -2 before and after the reaction.
Since there is a change in oxidation numbers for Cr and H, we can conclude that this equation describes a redox reaction. Cr is oxidized from +0.44 to +6, and H is oxidized from +1 to 0.
c. AgNO3 + KI → AgI + KNO3
- The oxidation number of Ag in AgNO3 can be calculated by using the fact that the sum of oxidation numbers is zero in a neutral compound. Given that O has an oxidation number of -2 and N has an oxidation number of +5 (as NO3 has a charge of -1), we can calculate it as follows:
x + 3(-2) + 5 = 0
x - 6 + 5 = 0
x - 1 = 0
x = 1
The oxidation number of Ag is +1 in AgNO3.
- The oxidation number of N in NO3 is +5 since O has an oxidation number of -2.
- The oxidation number of K is +1 since KI is a compound and the sum of the oxidation numbers is 0.
- The oxidation number of I in KI is -1.
- In AgI, Ag has an oxidation number of +1, and I has an oxidation number of -1.
- In KNO3, K still has an oxidation number of +1, and N has an oxidation number of +5, while O has an oxidation number of -2.
Let's analyze whether this equation describes a redox reaction:
- In Ag, the oxidation number is +1 both before and after the reaction.
- In N, the oxidation number is +5 before the reaction and +5 after the reaction.
- In K, the oxidation number is +1 both before and after the reaction.
- In I, the oxidation number is -1 both before and after the reaction.
Since there are no changes in oxidation numbers for any element, we can conclude that this equation does not describe a redox reaction.
d. 2PbS + 3O2 → 2PbO + 2SO2
- The oxidation number of Pb in PbS can be calculated by using the fact that the sum of oxidation numbers is zero in a neutral compound. Given that S has an oxidation number of -2, we can calculate it as follows:
2x + 2(-2) = 0
2x - 4 = 0
2x = 4
x = 2
The oxidation number of Pb is +2 in PbS.
- The oxidation number of S in PbS is -2.
- The oxidation number of O in O2 is 0.
- In PbO, Pb has an oxidation number of +2, and O has an oxidation number of -2.
- In SO2, S has an oxidation number of +4, and O has an oxidation number of -2.
Now, let's examine whether this equation describes a redox reaction:
- In Pb, the oxidation number is +2 both before and after the reaction.
- In S, the oxidation number is -2 before the reaction and +4 after the reaction.
- In O, the oxidation number is 0 before the reaction and -2 after the reaction.
Since there are changes in oxidation numbers for S and O, we can conclude that this equation describes a redox reaction. S is oxidized from -2 to +4, while O is reduced from 0 to -2.
Tell me what you don't understand about this problem.
Here is a very good site that gives you the rules for assigning oxidation states.
https://www.chemteam.info/Redox/Redox-Rules.html