lectrochemistry: Electrolysis
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Using the electrolytic process to produce two important gases:
When a concentrated aqueous solution of hydrochloric acid is electrolyzed, the positive hydrogen ions are attracted to the cathode, accept electrons from it, and are reduced to hydrogen gas.
2 e– + 2 H+(aq)
H2(g) cathode
The negative chloride ions are attracted to the anode, lose electrons, and are oxidized to chlorine gas.
2 Cl–(aq)
Cl2(g) + 2 e–
anode
The overall reaction is found by adding the electrode reactions.
2 H+(aq) + 2
Cl–(aq) H2(g) +
Cl2(g) net
This reaction, far from spontaneous, requires an external energy source. An important question might be, "How much of the gaseous products can we produce with a certain amount of current in a given period of time?". This becomes a simple stoichiometry problem once we know how many electrons are required for each reduction and how many are provided by the electric current (at sufficient voltage). Let's consider a specific problem so that we can see how a solution is determined:
Q: How long would it take to produce 36.0 grams of hydrogen gas in an electrolytic cell? Assume sufficient selective voltage and a current of 100.0 mA (milliamperes).
A: The reaction that produces hydrogen gas is the reduction that occurs at the cathode:
2 e– + 2
H+(aq) H2(g)
1 . How many moles do we need produced?
| solve from the number of given grams: | |||
| 36.0 g H2 | 1 mole H2 | = | 18.0 mole H2 |
| 2 g H2 | |||
2. How many moles of electrons do we need to reduce that many moles?
| solve from the balanced equation: | |||
| 18.0 mole H2 | 2 e– | = | 36.0 mole e– |
| 1 mole H2 | |||
3. How much charge is represented by this many electrons?
| solve from Faraday's constant: | |||
| 36.0 mole e– | 96,500 C | = | 3,470,000 C |
| 1 mole e– | |||
4. How much time is required to deliver this charge with a current of 100 mA?
| solve from amperage equation: | |||
| 3,470,000 C | 1 s | = | 34,700 s |
| 100.0 C | |||
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Electrolysis 
