Approach 1: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 e-
1621
Approach 1: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we use the
symbolic equation H2O + “O”
2 OH-(aq)
1622
Approach 1: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we use the
symbolic equation H2O + “O”
2 OH-(aq)
(use this for basic conditions)
1623
Approach 1: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we use the
symbolic equation H2O + “O”
2 OH-(aq)
(use this for basic conditions)
2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq)
1624
Approach 1: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we use the
symbolic equation H2O + “O”
2 OH-(aq)
(use this for basic conditions)
2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq)
With the charge balanced:
3 e- + 2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq)
1625
So the two balanced half-reactions are:
S2-(aq)
S(s) + 2 e-
(1)
1626
So the two balanced half-reactions are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq) (2)
1627
So the two balanced half-reactions are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq) (2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
1628
So the two balanced half-reactions are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq) (2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
1629
So the two balanced half-reactions are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq) (2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
6 e- + 4 H2O + 2 MnO4-(aq)
2 MnO2(s) + 8 OH-(aq) (2)
1630
So the two balanced half-reactions are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 2 H2O + MnO4-(aq)
MnO2(s) + 4 OH-(aq) (2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
6 e- + 4 H2O + 2 MnO4-(aq)
2 MnO2(s) + 8 OH-(aq) (2)
Now add:
3 S2-(aq) + 4 H2O +2 MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 8 OH-(aq)
1631
Approach 2: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
1632
Approach 2: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 e-
1633
Approach 2: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we treat it as
though the solution was acidic.
1634
Approach 2: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we treat it as
though the solution was acidic. Use the symbolic
equation: 2 H+(aq) + “O”
H2O
1635
Approach 2: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we treat it as
though the solution was acidic. Use the symbolic
equation: 2 H+(aq) + “O”
H2O
4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
1636
Approach 2: The skeleton equation is:
S2-(aq) + MnO4-(aq)
S(s) + MnO2(s)
The first half-equation is:
S2-(aq)
S(s) + 2 eTo balance the second half-equation, we treat it as
though the solution was acidic. Use the symbolic
equation: 2 H+(aq) + “O”
H2O
4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
With the charge balanced:
3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
1637
The two half-equations are:
1638
The two half-equations are:
S2-(aq)
S(s) + 2 e-
(1)
1639
The two half-equations are:
S2-(aq)
S(s) + 2 e3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
(1)
(2)
1640
The two half-equations are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
(2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
1641
The two half-equations are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
(2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
1642
The two half-equations are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
(2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
6e- + 8 H+(aq) + 2MnO4-(aq)
2MnO2(s) + 4 H2O (2)
1643
The two half-equations are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
(2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
6e- + 8 H+(aq) + 2MnO4-(aq)
2MnO2(s) + 4 H2O (2)
Now add the two half-equations:
1644
The two half-equations are:
S2-(aq)
S(s) + 2 e(1)
3 e- + 4 H+(aq) + MnO4-(aq)
MnO2(s) + 2 H2O
(2)
We need to find a common factor for the number of
electrons involved in both half-equations. That
factor is 6, so multiply the first reaction by 3 and the
second reaction by 2:
3 S2-(aq)
3 S(s) + 6 e(1)
6e- + 8 H+(aq) + 2MnO4-(aq)
2MnO2(s) + 4 H2O (2)
Now add the two half-equations:
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H 2O
1645
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
1646
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
1647
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
1648
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
(2) Do any cleanup that is necessary.
1649
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
(2) Do any cleanup that is necessary.
Add 8 OH- to both sides of the equation above.
1650
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
(2) Do any cleanup that is necessary.
Add 8 OH- to both sides of the equation above.
3 S2-(aq) + 8H2O + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4H2O + 8OH-(aq)
1651
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
(2) Do any cleanup that is necessary.
Add 8 OH- to both sides of the equation above.
3 S2-(aq) + 8H2O + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4H2O + 8OH-(aq)
Now cleanup the water:
1652
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
(2) Do any cleanup that is necessary.
Add 8 OH- to both sides of the equation above.
3 S2-(aq) + 8H2O + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4H2O + 8OH-(aq)
Now cleanup the water:
3 S2-(aq) + 4H2O + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 8OH-(aq)
1653
3 S2-(aq) + 8 H+(aq) + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4 H2O
Now we need to:
(1) Remove any H+ by using the reaction
H+ + OHH2O.
(2) Do any cleanup that is necessary.
Add 8 OH- to both sides of the equation above.
3 S2-(aq) + 8H2O + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 4H2O + 8OH-(aq)
Now cleanup the water:
3 S2-(aq) + 4H2O + 2MnO4-(aq)
3 S(s) + 2MnO2(s)
+ 8OH-(aq)
This matches the previous approach.
1654
Electrolysis
1655
Electrolysis
Electrolysis: A process that uses electrical energy
to drive a non-spontaneous reaction.
1656
Electrolysis
Electrolysis: A process that uses electrical energy
to drive a non-spontaneous reaction.
Electrolytic Cell: A device for carrying out
electrolysis.
1657
Electrolysis
Electrolysis: A process that uses electrical energy
to drive a non-spontaneous reaction.
Electrolytic Cell: A device for carrying out
electrolysis.
Electrochemical cell: A system that incorporates a
redox reaction to produce or use electrical energy.
1658
Electrode: The part of an electrochemical cell that
conducts electricity between the cell and the
surroundings.
1659
Electrode: The part of an electrochemical cell that
conducts electricity between the cell and the
surroundings.
Anode: Electrode where oxidation takes place.
1660
Electrode: The part of an electrochemical cell that
conducts electricity between the cell and the
surroundings.
Anode: Electrode where oxidation takes place.
Cathode: Electrode where reduction takes place.
1661
1662
1. Electrolysis of molten sodium chloride.
1663
1. Electrolysis of molten sodium chloride.
The subscript l indicates liquid (not aqueous) in the
following equations:
1664
1. Electrolysis of molten sodium chloride.
The subscript l indicates liquid (not aqueous) in the
following equations:
Cathode reaction: Na+(l) + eNa(l)
1665
1. Electrolysis of molten sodium chloride.
The subscript l indicates liquid (not aqueous) in the
following equations:
Cathode reaction: Na+(l) + eNa(l)
Anode reaction:
2 Cl-(l)
Cl2(g) + 2e-
1666
1. Electrolysis of molten sodium chloride.
The subscript l indicates liquid (not aqueous) in the
following equations:
Cathode reaction: Na+(l) + eNa(l)
Anode reaction:
2 Cl-(l)
Cl2(g) + 2eOverall reaction: 2Na+(l) + 2 Cl-(l)
2 Na(l) + Cl2(g)
1667
1. Electrolysis of molten sodium chloride.
The subscript l indicates liquid (not aqueous) in the
following equations:
Cathode reaction: Na+(l) + eNa(l)
Anode reaction:
2 Cl-(l)
Cl2(g) + 2eOverall reaction: 2Na+(l) + 2 Cl-(l)
2 Na(l) + Cl2(g)
In this example, electricity is being used to drive a
non-spontaneous reaction. Cl2(g) is an important
industrial chemical.
1668
1. Electrolysis of molten sodium chloride.
The subscript l indicates liquid (not aqueous) in the
following equations:
Cathode reaction: Na+(l) + eNa(l)
Anode reaction:
2 Cl-(l)
Cl2(g) + 2eOverall reaction: 2Na+(l) + 2 Cl-(l)
2 Na(l) + Cl2(g)
In this example, electricity is being used to drive a
non-spontaneous reaction. Cl2(g) is an important
industrial chemical.
The standard way to prepare reactive metals (such
as sodium) is by electrolysis.
1669
1670
2. Electrolysis of water (in the presence of a small
amount of H+(aq)).
Cathode reaction: 2 H+(aq) + 2 eH2(g)
Anode reaction: 2 H2O(l)
4 H+(aq) + O2(g) + 4eOverall reaction: 2 H2O(l)
2 H2(g) + O2(g)
1671
2. Electrolysis of water (in the presence of a small
amount of H+(aq)).
Cathode reaction: 2 H+(aq) + 2 eH2(g)
Anode reaction: 2 H2O(l)
4 H+(aq) + O2(g) + 4eOverall reaction: 2 H2O(l)
2 H2(g) + O2(g)
If electrolysis is attempted without the H+(aq)
present, nothing happens. There is no current flow
and no H2(g) or O2(g) is produced.
1672
2. Electrolysis of water (in the presence of a small
amount of H+(aq)).
Cathode reaction: 2 H+(aq) + 2 eH2(g)
Anode reaction: 2 H2O(l)
4 H+(aq) + O2(g) + 4eOverall reaction: 2 H2O(l)
2 H2(g) + O2(g)
If electrolysis is attempted without the H+(aq)
present, nothing happens. There is no current flow
and no H2(g) or O2(g) is produced. It is necessary
that the solution be an electrolyte.
1673
2. Electrolysis of water (in the presence of a small
amount of H+(aq)).
Cathode reaction: 2 H+(aq) + 2 eH2(g)
Anode reaction: 2 H2O(l)
4 H+(aq) + O2(g) + 4eOverall reaction: 2 H2O(l)
2 H2(g) + O2(g)
If electrolysis is attempted without the H+(aq)
present, nothing happens. There is no current flow
and no H2(g) or O2(g) is produced. It is necessary
that the solution be an electrolyte.
Electrolyte : A solution that conducts an electric
current due to the presence of ions.
1674
If the H+(aq) ions are provided by sulfuric acid for
example, both the cation and the anion play a role.
1675
If the H+(aq) ions are provided by sulfuric acid for
example, both the cation and the anion play a role.
In the vicinity of the anode there would be a build
up of positive charge (H+ ions are forming).
1676
If the H+(aq) ions are provided by sulfuric acid for
example, both the cation and the anion play a role.
In the vicinity of the anode there would be a build
up of positive charge (H+ ions are forming). It is
impossible for a region of solution to become
electrically charged.
1677
If the H+(aq) ions are provided by sulfuric acid for
example, both the cation and the anion play a role.
In the vicinity of the anode there would be a build
up of positive charge (H+ ions are forming). It is
impossible for a region of solution to become
electrically charged.
In the present case HSO4- and SO42- ions would
migrate to the anode to maintain electrical
neutrality of the solution.
1678
If the H+(aq) ions are provided by sulfuric acid for
example, both the cation and the anion play a role.
In the vicinity of the anode there would be a build
up of positive charge (H+ ions are forming). It is
impossible for a region of solution to become
electrically charged.
In the present case HSO4- and SO42- ions would
migrate to the anode to maintain electrical
neutrality of the solution. H+ ions would migrate to
the cathode region to prevent this region of the
solution be depleted of positive ions.
1679
3. Electrolysis of aqueous sodium chloride solution.
1680