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The electrolysis of aqueous sodium chloride (brine) is an important industrial process for the production of chlorine and sodium hydroxide. In fact, this process is the second largest consumer of electricity in the United States, after the production of aluminum. Write a balanced equation for the electrolysis of aqueous sodium chloride (hydrogen gas is also produced).

Short Answer

Expert verified
The balanced equation for the electrolysis of aqueous sodium chloride is: \(2Cl^-(aq) + 2H_2O(l) \rightarrow Cl_2(g) + H_2(g) + 2OH^-(aq)\).

Step by step solution

01

Identify the half-reactions

For the electrolysis of aqueous sodium chloride, two half-reactions occur - one at the anode where oxidation takes place and one at the cathode where reduction takes place. At the anode (the positively-charged electrode), chloride ions (Cl鈦) are oxidized to form chlorine gas (Cl鈧). The half-reaction is: Cl鈦(aq) 鈫 1/2 Cl鈧(g) + e鈦 At the cathode (the negatively-charged electrode), water molecules (H鈧侽) are reduced to form hydrogen gas (H鈧) and hydroxide ions (OH鈦): 2H鈧侽(l) + 2e鈦 鈫 H鈧(g) + 2OH鈦(aq)
02

Balance the half-reactions

The half-reactions above are already balanced in terms of mass and charge. Notice that each half-reaction has the appropriate number of electrons (e鈦) on each side of the reaction.
03

Combine the half-reactions

Now we need to combine the half-reactions to form the overall balanced equation for the electrolysis of aqueous sodium chloride. To do this, we must first make sure that the number of electrons in both half-reactions is equal. Currently, the anode half-reaction has 1 electron while the cathode half-reaction has 2 electrons. In order to balance the number of electrons, we will multiply the anode half-reaction by 2: 2Cl鈦(aq) 鈫 Cl鈧(g) + 2e鈦 Now, we can add the two half-reactions together: 2Cl鈦(aq) + 2H鈧侽(l) + 2e鈦 鈫 Cl鈧(g) + 2e鈦 + H鈧(g) + 2OH鈦(aq)
04

Simplify the equation

Lastly, simplify the equation by removing the electrons, as they are present on both sides of the equation: 2Cl鈦(aq) + 2H鈧侽(l) 鈫 Cl鈧(g) + H鈧(g) + 2OH鈦(aq) This is the balanced equation for the electrolysis of aqueous sodium chloride.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equations
Balanced chemical equations are essential to understanding chemical reactions. They ensure that the same number of each type of atom is present on both sides of the equation. In our context, we're focusing on the electrolysis of aqueous sodium chloride, which involves breaking down compounds into simpler substances using electricity.
In this process, sodium chloride (NaCl) in water is decomposed to produce chlorine gas and sodium hydroxide. A balanced equation for this process shows the substances involved at the beginning and end of the reaction, with equal numbers of atoms for each element.
When writing the balanced equation for the electrolysis of NaCl, the main goal is to ensure the number of electrons exchanged in each half-reaction is equal, leading to the overall reaction:
  • At the anode: 2Cl鈦(aq) 鈫 Cl鈧(g) + 2e鈦
  • At the cathode: 2H鈧侽(l) + 2e鈦 鈫 H鈧(g) + 2OH鈦(aq)
After canceling out electrons, we get:
  • 2Cl鈦(aq) + 2H鈧侽(l) 鈫 Cl鈧(g) + H鈧(g) + 2OH鈦(aq)
Industrial Chemical Processes
Electrolysis is one of the most significant industrial chemical processes. Specifically, the electrolysis of sodium chloride is critical for the production of chlorine and sodium hydroxide. These chemicals have vast applications in industries, including the manufacturing of plastics, paper, and detergents.
This process is energy-intensive and therefore consumes a lot of electricity, making it one of the major electricity-consuming processes worldwide. Despite its high energy consumption, its importance cannot be understated due to the pivotal role chlorine and sodium hydroxide play in various products and industries.
Mastering the chemical principles behind industrial processes like electrolysis can pave the way for innovations that improve energy efficiency and reduce costs.
Half-Reactions
In the context of electrolysis, understanding half-reactions is key to comprehending the overall chemical process. A half-reaction is one part of the overall reaction, either the oxidation or reduction process.
In electrolysis:
  • Oxidation occurs at the anode. For the aqueous sodium chloride electrolysis, the oxidation half-reaction is: Cl鈦(aq) 鈫 1/2 Cl鈧(g) + e鈦.
  • Reduction occurs at the cathode. Here, the water molecules are reduced, producing hydrogen gas and hydroxide ions: 2H鈧侽(l) + 2e鈦 鈫 H鈧(g) + 2OH鈦(aq).
Balancing these half-reactions in terms of mass and charge allows us to deduce the overall balanced chemical equation for the process.
Anode and Cathode Reactions
In electrolysis, reactions occur at two electrodes: the anode and the cathode. Understanding what happens at each is crucial for grasping the full process.
The **anode** is the positively charged electrode where oxidation occurs. For sodium chloride electrolysis, chloride ions lose electrons (get oxidized) to form chlorine gas. This can be represented as: 2Cl鈦(aq) 鈫 Cl鈧(g) + 2e鈦.
The **cathode** is the negatively charged electrode where reduction takes place. Water molecules gain electrons (get reduced), resulting in hydrogen gas and hydroxide ions: 2H鈧侽(l) + 2e鈦 鈫 H鈧(g) + 2OH鈦(aq).
The flow of electrons from the anode to the cathode drives these reactions, completing the circuit and facilitating the chemical transformations that produce valuable industrial chemicals.

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Most popular questions from this chapter

Many structures of phosphorus-containing compounds are drawn with some \(\mathrm{P}=\mathrm{O}\) bonds. These bonds are not the typical \(\pi\) bonds we've considered, which involve the overlap of two \(p\) orbitals. Instead, they result from the overlap of a \(d\) orbital on the phosphorus atom with a \(p\) orbital on oxygen. This type of \(\pi\) bonding is sometimes used as an explanation for why \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has the first structure below rather than the second: Draw a picture showing how a \(d\) orbital and a \(p\) orbital overlap to form a \(\pi\) bond.

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