Complete and balance the following equations, and identify the oxidizing and red

Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Use the lowest possible coefficients. Include states-of-matter (s), (l), (g), (aq) similar to the given reaction.)

a) H2O2(aq) + Cl2O7(aq) ? ClO2-(aq) + O2(g) (basic solution) oxidizing and reducing agent?

b)   MnO4-(aq) + CH3OH(aq) ? Mn2+(aq) + HCO2H(aq) (acidic solution) oxidizing and reducing agent?

c) As2O3(s) + NO3-(aq) ? H3AsO4(aq) + N2O3(aq) (acidic solution) oxidizing and reducing agent?

d) MnO4-(aq) + Cl-(aq) ? Mn2+(aq) + Cl2(g) (acidic solution) oxidizing and reducing agent?

Explanation / Answer

a) H2O2(aq) + Cl2O7(aq) ---------------> ClO2-(aq) + O2(g) (basic solution) oxidizing and reducing agent?

Oxidation half-reaction:

H2O2(aq) ---> O2(g)

Here, we need to balance the Hydrogen atoms as Oxygen atoms are already balanced.

H2O2(aq) ---> O2(g) + 2H+

Now, inorder to balnce the charges,

H2O2(aq) ---> O2(g) + 2H+ + 2e- ------- (1)

Reduction half-reaction:

Cl2O7(aq) ---------------> ClO2-(aq)

First we need to balance Cl atoms

Cl2O7(aq) --------> 2ClO2^-(aq)

Now, we need to balance Oxygen atoms

Cl2O7(aq) -----------> 2ClO2^-(aq) + 3H2O

Now, hydrogen atoms are balanced.

Cl2O7(aq) + 6H+ ----------> 2ClO2^-(aq) + 3H2O

Finally charges are balanced.

Cl2O7(aq) + 6H+ + 8e- -------------> 2ClO2^-(aq) + 3H2O ----------- (2)

Eq(1) x 4 + Eq(2)

4H2O2(aq) -------------> 4O2(g) + 8H+ + 8e-

Cl2O7(aq) + 6H+ + 8e- -------------> 2ClO2^-(aq) + 3H2O

-------------------------------------------------------------------------------------------------

4H2O2(aq) + Cl2O7(aq) -------------> 4O2(g) + 2H+ + 2ClO2^-(aq) + 3H2O

Adding 2OH- on both sides as it is basic solution,

4H2O2(aq) + Cl2O7(aq) + 2OH- -------------> 4O2(g) + 2ClO2^-(aq) + 5H2O

Thus, the balanced equation is

4H2O2(aq) + Cl2O7(aq) + 2OH- -------------> 4O2(g) + 2ClO2^-(aq) + 5H2O

Oxidizing agent : Cl2O7

Reducing agent : H2O2

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b)   MnO4-(aq) + CH3OH(aq) -----------------> Mn2+(aq) + HCO2H(aq) (acidic solution) oxidizing and reducing agent?

Oxidation half-reaction:

CH3OH(aq) -----------------> HCOOH(aq)

Here, we need to balance the Oxygen atoms.

CH3OH(aq) + H2O -----------------> HCOOH(aq)

Now we need to balance Hydrogen atoms.

CH3OH(aq) + H2O -----------------> HCOOH(aq) + 4H+

Now, inorder to balnce the charges,

CH3OH(aq) + H2O -----------------> HCOOH(aq) + 4H+ + 4e- ------- (1)

Reduction half-reaction:

MnO4-(aq) -----------------> Mn2+(aq)

First we need to balance Oxygen atoms

MnO4-(aq) -----------------> Mn2+(aq) + 4H2O

We need to balance the hydrogen atoms

MnO4-(aq) + 8H+ -----------------> Mn2+(aq) + 4H2O

Now, we need to balance the electrons.

MnO4-(aq) + 8H+ + 5e- -----------------> Mn2+(aq) + 4H2O ----------- (2)

Eq(1) x 5 + Eq(2) x 4

5CH3OH(aq) + 5H2O -----------------> 5HCOOH(aq) + 20H+ + 20e-

4MnO4-(aq) + 32H+ + 20e- -----------------> 4Mn2+(aq) + 16H2O

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5CH3OH(aq) + 4MnO4-(aq) + 12H+(aq) -----------------> 5HCOOH(aq) + 4Mn2+(aq) + 11H2O(l)

Thus, the balanced equation is

5CH3OH(aq) + 4MnO4-(aq) + 12H+(aq) -----------------> 5HCOOH(aq) + 4Mn2+(aq) + 11H2O(l)

Oxidizing agent : MnO4-

Reducing agent : CH3OH

--------------------------------------------------------------------------------------------------------------------------------

c) As2O3(s) + NO3-(aq) ---------------------> H3AsO4(aq) + N2O3(aq) (acidic solution) oxidizing and reducing agent?

Oxidation half-reaction:

As2O3(s) ---------------------> H3AsO4(aq)

First we need to balance the As atoms.

As2O3(s) ---------------------> 2H3AsO4(aq)

Here, we need to balance the Oxygen atoms.

As2O3(s) + 5H2O ---------------------> 2H3AsO4(aq)

Now we need to balance Hydrogen atoms.

As2O3(s) + 5H2O ---------------------> 2H3AsO4(aq) + 4H+

Now, inorder to balnce the charges,

As2O3(s) + 5H2O ---------------------> 2H3AsO4(aq) + 4H+ + 4e- ------- (1)

Reduction half-reaction:

NO3-(aq) ---------------------> N2O3(aq)

First we need to balance Nitrogen atoms

2NO3-(aq) ---------------------> N2O3(aq)

We need to balance the oxygen atoms

2NO3-(aq) ---------------------> N2O3(aq) + 3H2O

We need to balance the hydrogen atoms

2NO3-(aq) + 6H+ ---------------------> N2O3(aq) + 3H2O

Now, we need to balance the electrons.

2NO3-(aq) + 6H+ + 4e- ---------------------> N2O3(aq) + 3H2O ----------- (2)

Eq(1) + Eq(2)

As2O3(s) + 5H2O ---------------------> 2H3AsO4(aq) + 4H+ + 4e-

2NO3-(aq) + 6H+ + 4e- ---------------------> N2O3(aq) + 3H2O

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As2O3(s) + 2H2O(l) + 2NO3-(aq) + 2H+(aq) ---------------------> 2H3AsO4(aq) + N2O3(aq)

Thus, the balanced equation is

As2O3(s) + 2H2O(l) + 2NO3-(aq) + 2H+(aq) ---------------------> 2H3AsO4(aq) + N2O3(aq)

Oxidizing agent : NO3-

Reducing agent : As2O3

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d) MnO4-(aq) + Cl-(aq) -----------> Mn2+(aq) + Cl2(g) (acidic solution) oxidizing and reducing agent?

Oxidation half-reaction:

Cl-(aq) -----------> Cl2(g)

Here, we need to balance the Chlorine atoms.

2Cl-(aq) -----------> Cl2(g)

Now, inorder to balnce the charges,

2Cl-(aq) -----------> Cl2(g) + 2e- ------- (1)

Reduction half-reaction:

MnO4-(aq) -----------------> Mn2+(aq)

First we need to balance Oxygen atoms

MnO4-(aq) -----------------> Mn2+(aq) + 4H2O

We need to balance the hydrogen atoms

MnO4-(aq) + 8H+ -----------------> Mn2+(aq) + 4H2O

Now, we need to balance the electrons.

MnO4-(aq) + 8H+ + 5e- -----------------> Mn2+(aq) + 4H2O ----------- (2)

Eq(1) x 5 + Eq(2) x 2

10Cl-(aq) -----------> 5Cl2(g) + 10e-

2MnO4-(aq) + 16H+ + 10e- -----------------> 2Mn2+(aq) + 8H2O  

-------------------------------------------------------------------------------------------------

10Cl-(aq) + 2MnO4-(aq) + 16H+(aq)   -----------> 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)

Thus, the balanced equation is

10Cl-(aq) + 2MnO4-(aq) + 16H+(aq)   -----------> 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)

Oxidizing agent : MnO4-

Reducing agent : Cl-

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