Question

Consider the electrolysis of \(\mathrm{CuCl}_{2}\) to form \(\mathrm{Cu}(s)\) and \(\mathrm{Cl}_{2}(\mathrm{~g}) .\) Calculate the minimum voltage required to carry out this reaction at standard conditions. If a voltage of \(1.50 \mathrm{~V}\) is actually used, how many kilojoules of electrical energy are consumed in producing \(2.00 \mathrm{~g}\) of \(\mathrm{Cu} ?\)

Short answer

Answer: The minimum voltage required for the electrolysis of CuCl2 is 1.02V, and 4.5582 kJ of electrical energy is consumed to produce 2g of Cu when a voltage of 1.50V is used.

Step-by-step solution

Write the half-reactions

For the electrolysis of CuCl2, the half-reactions are: Oxidation: \(2\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g})+2e^{-}\) Reduction: \(\mathrm{Cu}^{2+}(\mathrm{aq})+2e^{-} \rightarrow \mathrm{Cu}(s)\)

Calculate the standard cell potential

Standard electrode potentials for the reactions are: Oxidation: \(E^{o}_{\mathrm{Cl}^{-}/\mathrm{Cl}_{2}} = +1.36\,\mathrm{V}\) Reduction: \(E^{o}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34\,\mathrm{V}\) Calculate the standard cell potential \(\Delta E^{o}\): \(\Delta E^{o} = E^{o}_{\mathrm{oxidation}} - E^{o}_{\mathrm{reduction}}\) \(\Delta E^{o} = (+1.36\,\mathrm{V}) - (+0.34\,\mathrm{V})\) \(\Delta E^{o} = +1.02\,\mathrm{V}\) The minimum voltage required for the electrolysis of CuCl2 is \(1.02\,\mathrm{V}\).

Calculate the moles of Cu produced

Given the mass of Cu to be produced (2g), determine the number of moles: \(n_{\mathrm{Cu}} = \frac{2\,\mathrm{g}}{63.55\frac{\mathrm{g}}{\mathrm{mol}}}=0.0315\,\mathrm{mol}\)

Calculate the charge needed for the electrolysis

\(dq = n_{\mathrm{Cu}} \times F\) Where \(dq\) is the charge (Coulombs), \(n_{\mathrm{Cu}}\) is the moles of copper produced, and \(F\) is Faraday's constant, \(9.6485\times10^{4}\frac{\mathrm{C}}{\mathrm{mol}}\). \(dq = 0.0315\,\mathrm{mol} \times 9.6485\times10^{4}\frac{\mathrm{C}}{\mathrm{mol}}=3.0388\times10^3\,\mathrm{C}\)

Calculate the electrical energy consumed

Use the formula: \(W = V \times dq\) Where \(W\) is the work done (Joules), \(V\) is the voltage (Volts), and \(dq\) is the charge (Coulombs). \(W = 1.50\,\mathrm{V} \times 3.0388\times10^3\,\mathrm{C} = 4.5582\times10^3\,\mathrm{J}\) Convert the energy to kilojoules: \(W = 4.5582\,\mathrm{kJ}\) Thus, \(4.5582\,\mathrm{kJ}\) of electrical energy is consumed in producing 2g of Cu when a voltage of 1.50V is used.

Key concepts

Electrode Potential

In electrochemistry, electrode potential is a crucial concept that helps determine the driving force behind redox reactions in electrolytic cells.\(\ E^{o}\) (standard electrode potential) is a measure of the tendency of a chemical species to gain or lose electrons and become reduced or oxidized. For our example involving \(\mathrm{CuCl}_{2}\)

• The oxidation half-reaction involves chloride ions \(\, \mathrm{Cl}^{-} \), transitioning to chlorine gas \(\, \mathrm{Cl}_{2} \).
• The reduction half-reaction takes place when copper ions \(\, \mathrm{Cu}^{2+} \) gain electrons forming solid copper \(\, \mathrm{Cu} \).

Standard electrode potentials for these reactions are known from tables of standard reduction potentials. Calculating the standard cell potential \( \Delta E^{o} \) involves subtracting the sum of reduction potentials:\[\Delta E^{o} = E^{o}_{\mathrm{oxidation}} - E^{o}_{\mathrm{reduction}}\]For the given reactions, this potential tells us the minimum voltage required for the electrolytic reaction to occur, which is found to be \(\, 1.02 \, \mathrm{V} \). This means that any voltage over \( 1.02 \, \mathrm{V} \) will drive the electrolysis reaction forward.

Faraday's Constant

Faraday's Constant \( (F) \) plays a vital role in connecting moles of electrons to the charge passed in electrochemical reactions. It's essentially the magnitude of electric charge per mole of electronsand has a constant value of \( 9.6485 \times 10^4 \, \mathrm{C/mol} \).In electrolysis, knowing how much of a substance can be transformed helps in quantifying & calculating the exact charge needed. To calculate charge, we can use the formula:\[q = n \times F\]where \( q \) is the charge in coulombs, and \( n \) represents moles of electrons exchanged during the reaction.For example: If producing \(\, 2 \, \mathrm{g} \, of \, Cu \), determines \( 0.0315 \, \mathrm{mol} \).Multiplying this by Faraday's Constant gives us the total charge needed to affect the oxidation/reduction reactions.This relation effectively helps translate the solute’s change in massto the charge employed, driving forward the electrolytic procedure.Understanding this fundamental idea is critical when considering reactionsand their real-life applications in devices like batteries or in engineering electroplating processes.

Electrical Energy Calculation

To determine the electrical energy consumed during electrolysis, we calculate using the voltage applied and the total charge.Energy represents the work done to force the reaction to go forward,and we use this straightforward relationship: \[ W = V \times q \]where \( W \) is the work (or energy) in Joules,\( V \) represents voltage in volts, and \( q \) is the charge in coulombs.For our copper electrolysis example, an applied voltage of \( 1.50 \, \mathrm{V} \) and total chargecalculated from moles of \(\, 2 \, \mathrm{g} \, of \, Cu \) translate to \(\, 4.5582 \, \mathrm{kJ} \).This tells us how much energy is consumed to produce \(\, 2 \, \mathrm{g} \, of \, \mathrm{Cu} \) from our reaction setup attributed to the given conditions.Understanding this concept is crucial, especially when considering commercial or industrial scale operations where energy efficiency and resource optimization are key to sustainable practices.Moreover, these calculations help in setting parameters and understanding the intricacies of electrochemical setups.