Balance the following equations:
1. MnO4- + C2O42- = Mn2+ + CO2
2. NO2- + MnO4- = NO3- + Mn2+ (in acid Solution)
3. I- + MnO4- = I2 = MnO2 (In basic Solution)
4. Cr2O72- + Fe2+ = Cr3+ + Fe3+
5. MnO4- + Fe2+ = Mn2+ + Fe3+
6. Bi (OH)3 + SnO22- = SnO32- + Bi (Basic Solution)
7. P + Cu2+ = Cu + H2PO4-
Oxidation Numbers
1. Determine the oxidation number of the atom accompanying the oxygen in:
a. SO3
b. NO3 –
c. N2O5
d. MnO4 –
e. ClO3 –
f. CrO4 2-
g. Cr2O7 2-
h. PO4 3-
i. SO3 2-
2. Determine the oxidation number of the nitrogen in:
a. N2
b. NH3
c. N2H4
d. NO
e. NF3
f. KNO2
g. (NH4)2SO4
h. NH4+
i. N2O
3. What is the oxidation number for each of the following?
a. sulphur in Na2S2O5?
b. phosphorus in K2HPO4?
c. chromium in K2Cr2O7?
Balancing Redox Equations:
1. MnO4- + C2O42- = Mn2+ + CO2
Balanced Equation: 2 MnO4^- + 5 C2O4^2- + 16 H+ → 2 Mn^2+ + 10 CO2 + 8 H2O
2. NO2- + MnO4- = NO3- + Mn2+ (in acid solution)
Balanced Equation: 3 NO2^- + MnO4^- + 4 H+ → 3 NO3^- + Mn^2+ + 2 H2O
3. I- + MnO4- = I2 = MnO2 (In basic solution)
Balanced Equation: 2 I^- + MnO4^- + 2 H2O → I2 + MnO2 + 4 OH^-
4. Cr2O72- + Fe2+ = Cr3+ + Fe3+
Balanced Equation: Cr2O7^2- + 6 Fe^2+ + 14 H+ → 2 Cr^3+ + 6 Fe^3+ + 7 H2O
5. MnO4- + Fe2+ = Mn2+ + Fe3+
Balanced Equation: MnO4^- + 5 Fe^2+ + 8 H+ → Mn^2+ + 5 Fe^3+ + 4 H2O
6. Bi(OH)3 + SnO22- = SnO32- + Bi (Basic Solution)
Balanced Equation: Bi(OH)3 + 3 SnO2^2- → 3 SnO3^2- + Bi + 3 OH^-
7. P + Cu2+ = Cu + H2PO4-
Balanced Equation: 3 P + 2 Cu^2+ + 6 H2O → 3 H2PO4^- + 2 Cu + 6 H+
Oxidation Numbers:
1. Determine the oxidation number of the atom accompanying the oxygen in:
a. SO3 → Sulfur (S) has an oxidation number of +6.
b. NO3– → Nitrogen (N) has an oxidation number of +5.
c. N2O5 → Nitrogen (N) has an oxidation number of +5.
d. MnO4– → Manganese (Mn) has an oxidation number of +7.
e. ClO3– → Chlorine (Cl) has an oxidation number of +5.
f. CrO4 2- → Chromium (Cr) has an oxidation number of +6.
g. Cr2O7 2- → Chromium (Cr) has an oxidation number of +6.
h. PO4 3- → Phosphorus (P) has an oxidation number of +5.
i. SO3 2- → Sulfur (S) has an oxidation number of +4.
2. Determine the oxidation number of the nitrogen in:
a. N2 → Nitrogen (N) has an oxidation number of 0.
b. NH3 → Nitrogen (N) has an oxidation number of -3.
c. N2H4 → Nitrogen (N) has an oxidation number of -2.
d. NO → Nitrogen (N) has an oxidation number of +2.
e. NF3 → Nitrogen (N) has an oxidation number of +3.
f. KNO2 → Nitrogen (N) has an oxidation number of +3.
g. (NH4)2SO4 → Nitrogen (N) has an oxidation number of -3.
h. NH4+ → Nitrogen (N) has an oxidation number of -3.
i. N2O → Nitrogen (N) has an oxidation number of +1.
3. What is the oxidation number for each of the following?
a. Sulphur in Na2S2O5 → Sulfur (S) has an oxidation number of +2.5 on average. (Here, there are two S with different oxidation states: +6 and -2, respectively).
b. Phosphorus in K2HPO4 → Phosphorus (P) has an oxidation number of +5.
c. Chromium in K2Cr2O7 → Chromium (Cr) has an oxidation number of +6.
3. What is the oxidation number for each of the following?
a. sulphur in Na2S2O5^?
b. phosphorus in K2HPO4^?
c. chromium in K2Cr2O7^?
To determine the oxidation numbers of the specified elements in the given compounds, let's analyze the compounds one by one:
a. Sulfur in Na2S2O5 (Sodium metabisulfite)
In Na2S2O5, the oxidation state of sodium (Na) is +1. For the sulfate ion part (S2O5)^2-, the overall charge is -2. Here, we have one sulfur with an oxidation state typical of a sulfate ion (SO4)^2- which would be +6, and the other sulfur is in a lower oxidation state typical of a sulfite ion (SO3)^2- which would be +4. The average oxidation state of sulfur in this compound doesn't accurately reflect the individual oxidation states since there are two distinct sulfur atoms. To find the individual states we use the oxygen having an oxidation number of -2 and calculate as follows:
For the SO4-like part:
S(oxidation state) + 4(-2) = -2 (charge of the ion)
S(oxidation state) = +6
For the SO3-like part (assuming it loses an oxygen to the SO4):
S(oxidation state) + 3(-2) = +2 (since it is linked to the sulfite-like sulfur)
S(oxidation state) = +5
That said, in Na2S2O5, one sulfur is in a +5 oxidation state, and the other is in +6.
b. Phosphorus in K2HPO4 (Dipotassium hydrogen phosphate)
In K2HPO4, potassium (K) has an oxidation state of +1, and the overall charge of the ion (HPO4)^2- is -2. Since hydrogen has an oxidation state of +1 and oxygen has an oxidation state of -2, the calculation for the phosphorus oxidation number is:
1(+1) + P(oxidation state) + 4(-2) = -2
+1 + P(oxidation state) -8 = -2
P(oxidation state) = -2 + 8 - 1
P(oxidation state) = +5
The oxidation state of phosphorus is +5.
c. Chromium in K2Cr2O7 (Potassium dichromate)
In K2Cr2O7, potassium (K) has an oxidation state of +1. Since there are two K atoms, their total contribution is +2. The overall charge of the compound is neutral (0). Oxygen (O) always has an oxidation states of -2 unless in peroxides or bound to fluorine. The calculation for the chromium oxidation number is:
2(+1) + 2(Cr(oxidation state)) + 7(-2) = 0
2 + 2(Cr(oxidation state)) -14 = 0
2(Cr(oxidation state)) = 14 - 2
2(Cr(oxidation state)) = 12
Cr(oxidation state) = +6
Therefore, the oxidation state of chromium in K2Cr2O7 is +6.
3. What is the oxidation number for each of the following?
a. sulphur in Na2S2O5?
b. phosphorus in K2HPO4?
c. chromium in K2Cr2O7?
To calculate the oxidation numbers for sulfur, phosphorus, and chromium in the respective compounds given (Na2S2O5, K2HPO4, and K2Cr2O7), we'll use the rules for oxidation numbers along with the known charges of other elements and ions within those compounds.
a. Sulfur in Na2S2O5 (Sodium metabisulfite)
The overall charge of the compound is neutral. The oxidation number for sodium (Na) is +1, and for oxygen (O) is -2. Because we have two Na atoms, we have a total of +2 from sodium. There are five O atoms, each with an oxidation number of -2, for a total of -10 from oxygen.
Let the average oxidation number of the two sulfur atoms be x. Therefore, we can set up the following equation using the sum of oxidation numbers in a neutral compound which is equal to zero:
2(+1) + 2x + 5(-2) = 0
2 + 2x - 10 = 0
2x - 8 = 0
2x = 8
x = 4
However, it is important to note that in Na2S2O5, there are two sulfurs with different oxidation states due to the bisulfite ion. If one sulfur has an oxidation number of +5 (typical of a sulfite ion, S(+4) and an additional oxidation), and the other must have an oxidation state less than +4 to balance the total +8 found. Therefore, one sulfur will have an oxidation number of +5, and the other will have an oxidation number of +3. This totals +8 which is the sum needed for the two sulfur atoms together.
b. Phosphorus in K2HPO4 (Dipotassium hydrogen phosphate)
For the compound K2HPO4, potassium (K) has an oxidation number of +1 and we have two K atoms. Hydrogen (H) typically has an oxidation number of +1 and oxygen (O) -2. Let the oxidation number of phosphorus (P) in this compound be y.
We can set up the following equation for the polyatomic ion (HPO4)^2- because the overall charge of the compound is neutral:
2 + 1 + y + 4(-2) = 0
3 + y - 8 = 0
y - 5 = 0
y = +5
So, the oxidation number of phosphorus in K2HPO4 is +5.
c. Chromium in K2Cr2O7 (Potassium dichromate)
For the compound K2Cr2O7, potassium (K) has an oxidation number of +1 and we have two K atoms. Oxygen (O) has an oxidation number of -2. Let the oxidation number of chromium (Cr) in this compound be z.
2(+1) + 2z + 7(-2) = 0
2 + 2z - 14 = 0
2z - 12 = 0
2z = 12
z = +6
So, the oxidation number of chromium in K2Cr2O7 is +6. There are two chromium atoms in the compound, each with an oxidation number of +6.
To balance the equations provided, you need to apply the concept of balancing redox reactions. Here's how you can balance each of the equations:
1. MnO4- + C2O42- = Mn2+ + CO2:
- Assign oxidation numbers:
Mn in MnO4- is +7, C in C2O42- is +3.
- Identify the changes in oxidation numbers:
Mn goes from +7 to +2 (reduction), C goes from +3 to +4 (oxidation).
- Balance the equation:
Add coefficients to balance the number of atoms:
2MnO4- + 5C2O42- = 2Mn2+ + 10CO2
2. NO2- + MnO4- = NO3- + Mn2+ (in acid solution):
- Assign oxidation numbers:
N in NO2- is -3, O in MnO4- is -2.
- Identify the changes in oxidation numbers:
N goes from -3 to +5 (oxidation), Mn goes from +7 to +2 (reduction).
- Balance the equation:
Add coefficients to balance the number of atoms:
5NO2- + 2MnO4- + 6H+ = 5NO3- + 2Mn2+ + 3H2O
3. I- + MnO4- = I2 + MnO2 (in basic solution):
- Assign oxidation numbers:
I in I- is -1, O in MnO4- is -2.
- Identify the changes in oxidation numbers:
I goes from -1 to 0 (oxidation), Mn goes from +7 to +4 (reduction).
- Balance the equation:
Add coefficients to balance the number of atoms:
8OH- + 2MnO4- + 10I- = 5I2 + 2MnO2 + 4H2O
4. Cr2O72- + Fe2+ = Cr3+ + Fe3+:
- Assign oxidation numbers:
Cr in Cr2O72- is +6, Fe in Fe2+ is +2.
- Identify the changes in oxidation numbers:
Cr goes from +6 to +3 (reduction), Fe goes from +2 to +3 (oxidation).
- Balance the equation:
Add coefficients to balance the number of atoms:
6H+ + Cr2O72- + 6Fe2+ = 2Cr3+ + 6Fe3+ + 7H2O
5. MnO4- + Fe2+ = Mn2+ + Fe3+:
- Assign oxidation numbers:
Mn in MnO4- is +7, Fe in Fe2+ is +2.
- Identify the changes in oxidation numbers:
Mn goes from +7 to +2 (reduction), Fe goes from +2 to +3 (oxidation).
- Balance the equation:
Add coefficients to balance the number of atoms:
8H+ + MnO4- + 5Fe2+ = Mn2+ + 5Fe3+ + 4H2O
6. Bi(OH)3 + SnO22- = SnO32- + Bi (basic solution):
- This equation is already balanced.
7. P + Cu2+ = Cu + H2PO4-:
- Assign oxidation numbers:
P in P is 0, Cu in Cu2+ is +2.
- Identify the changes in oxidation numbers:
P goes from 0 to +5 (oxidation), Cu goes from +2 to 0 (reduction).
- Balance the equation:
Add coefficients to balance the number of atoms:
3Cu2+ + 2P + 6H2O = 3Cu + 2H2PO4-
Now, onto determining the oxidation numbers:
1. Oxidation numbers in compounds:
a. SO3: S is +6, O is -2
b. NO3-: N is +5, O is -2
c. N2O5: N is +5, O is -2
d. MnO4-: Mn is +7, O is -2
e. ClO3-: Cl is +5, O is -2
f. CrO4 2-: Cr is +6, O is -2
g. Cr2O7 2-: Cr is +6, O is -2
h. PO4 3-: P is +5, O is -2
i. SO3 2-: S is +4, O is -2
2. Oxidation numbers in nitrogen compounds:
a. N2: N is 0
b. NH3: N is -3
c. N2H4: N is 0
d. NO: N is +2
e. NF3: N is +3
f. KNO2: N is +3
g. (NH4)2SO4: N is -3
h. NH4+: N is -3
i. N2O: N is +1
3. Oxidation number in specific compounds:
a. sulphur in Na2S2O5: S is +4
b. phosphorus in K2HPO4: P is +5
c. chromium in K2Cr2O7: Cr is +6
Remember, oxidation numbers are assigned based on a set of rules, and they represent the charge an atom would have in a compound if electrons were strictly assigned.