Problem 50 - Chemistry

Question

In the electrolysis of an aqueous \(\mathrm{AgNO}_{3}\) solution, \(0.67 \mathrm{~g}\) of Ag is deposited after a certain period of time. (a) Write the half-reaction for the reduction of \(\mathrm{Ag}^{+} .\) (b) What is the probable oxidation half-reaction? (c) Calculate the quantity of electricity used (in coulombs).

Step-by-Step Solution

Answer

The reduction half-reaction is \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \). Probable oxidation is water to \(\mathrm{O}_2\). Electricity used: 599.8 C.

Step 1: Identify the Reduction Half-Reaction

In electrolysis, silver ions \( \mathrm{Ag}^{+}\) in solution are reduced to form solid silver (\(\mathrm{Ag}\)). The half-reaction for this reduction can be written as: \[ \mathrm{Ag}^{+} (aq) + e^{-} \rightarrow \mathrm{Ag} (s) \] This equation shows that one mole of electrons is required to reduce one mole of silver ions.

Step 2: Determine the Probable Oxidation Half-Reaction

In the electrolysis of an aqueous solution, water can be oxidized at the anode. The common oxidation half-reaction for water in acidic or neutral solution is: \[ 2\mathrm{H}_2\mathrm{O} (l) \rightarrow \mathrm{O}_2 (g) + 4\mathrm{H}^{+} (aq) + 4e^{-} \] Thus, water is the probable species undergoing oxidation.

Step 3: Calculate Moles of Ag Deposited

To find the moles of silver deposited, use the molar mass of silver (\(107.87 \ \mathrm{g/mol}\)):\[ \text{Moles of } \mathrm{Ag} = \frac{0.67 \ \mathrm{g}}{107.87 \ \mathrm{g/mol}} \approx 0.006215 \ \text{mol} \] This tells us how many moles of \(\mathrm{Ag}^{+}\) were reduced.

Step 4: Calculate Faradays of Electricity Used

One mole of \(\mathrm{Ag}^{+}\) requires one mole of electrons (or one Faraday) for reduction. Therefore, the moles of electrons used is the same as the moles of \(\mathrm{Ag}\):\[ 0.006215 \ \text{mol of electrons (or Faradays)} \]

Step 5: Convert Faradays to Coulombs

Since 1 Faraday = 96485 coulombs, we convert the moles of electrons to coulombs:\[ \text{Coulombs} = 0.006215 \times 96485 \ \mathrm{C/mol} \approx 599.8 \ \mathrm{C} \] This is the total quantity of electricity used in the electrolysis process.

Key Concepts

Reduction Half-Reaction

In electrolysis, one of the key processes involves the reduction at the cathode. This is where the reduction half-reaction occurs. For the aqueous solution of silver nitrate ([\text{AgNO}_3]), silver ions ([\text{Ag}^+]) in the solution gain electrons to form solid silver. This can be expressed by the equation:\[\text{Ag}^+ (aq) + e^- \rightarrow \text{Ag} (s)\]This simple equation illustrates that one silver ion requires one electron to be reduced into one atom of solid silver. Understanding the reduction half-reaction is key because it tells us how many electrons are involved in the process, which directly links to the amount of electricity used in the process.
Reduction half-reactions are crucial in processes like electroplating, refining metals, and operating batteries, where the transfer of electrons results in the desired metal deposition or reduction of ions.

Oxidation Half-Reaction

Parallel to the process of reduction at the cathode is oxidation happening at the anode. For the given aqueous [\text{AgNO}_3] solution, the probable oxidation half-reaction involves water, rather than nitrate ions. This oxidation can be represented by the equation:\[2\text{H}_2\text{O} (l) \rightarrow \text{O}_2 (g) + 4\text{H}^+ (aq) + 4e^-\]Here, four electrons are released as water is oxidized to oxygen gas and hydrogen ions, freeing up the electrons needed for the reduction half-reaction on the cathode side.

• The choice of this reaction instead of others is due to its lower energy requirements.
• This makes it a more likely candidate in neutral or acidic solutions.

Understanding oxidation half-reactions helps in predicting the by-products of electrolysis and enhances knowledge of how current flows in electrochemical cells.

Faraday's Law of Electrolysis

Faraday's Law of Electrolysis is a central principle in understanding how much substance can be transformed in an electrolysis reaction. The law states that the amount of chemical change is proportional to the quantity of electricity that passes through the electrolyte. In simpler terms:
"\[\text{mass of substance deposited or evolved}]= \text{electrons transferred in moles} \times \text{Faraday's constant}\]For the case with silver deposition, this was calculated using the known molar mass of silver and the measured mass deposited to find moles of electrons transferred.

• [\text{1 Faraday} = 96485 \text{ C/mol}], which provides a direct conversion from moles of electrons to total charge in coulombs.

Understanding this law allows you to calculate not only how much charge has been used but also precisely how much of a material will be deposited or dissolved during electrolysis.

Molar Mass Calculation

Calculating the molar mass of a substance is an essential step in many chemistry calculations, including those involved in electrolysis. In simple words, molar mass relates the weight of a sample to its amount in moles. It gives the weight of one mole of a given element or compound:
"\[\text{Molar Mass} = \frac{\text{mass in grams}}{\text{amount in moles}}\]For silver in the given electrolysis problem:\[\text{Molar Mass of Silver} \approx 107.87 \text{ g/mol}\]Using this, along with the mass of silver deposited, allows the calculation of moles and, consequently, the determination of electrons involved through stoichiometry. Accurate molar mass calculation is key in translating grams into moles, which is the bridge to all stoichiometric predictions in chemical reactions, including electrolysis.