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Chapter 18: Problem 78

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

Expert verified
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

01

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: \(\mathrm{Cu}^{2+}(\mathrm{aq})+2e^{-} \rightarrow \mathrm{Cu}(s)\)
02

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 \(1.02\,\mathrm{V}\).
03

Calculate the moles of Cu produced

Given the mass of Cu to be produced (2g), determine the number of moles: \(n_{\mathrm{Cu}} = \frac{2\,\mathrm{g}}{63.55\frac{\mathrm{g}}{\mathrm{mol}}}=0.0315\,\mathrm{mol}\)
04

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, \(9.6485\times10^{4}\frac{\mathrm{C}}{\mathrm{mol}}\). \(dq = 0.0315\,\mathrm{mol} \times 9.6485\times10^{4}\frac{\mathrm{C}}{\mathrm{mol}}=3.0388\times10^3\,\mathrm{C}\)
05

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.

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Most popular questions from this chapter

Consider the cell $$ \mathrm{Pt}\left|\mathrm{H}_{2}\right| \mathrm{H}^{+} \| \mathrm{H}^{+}\left|\mathrm{H}_{2}\right| \mathrm{Pt} $$ In the anode half-cell, hydrogen gas at \(1.0 \mathrm{~atm}\) is bubbled over a platinum electrode dipping into a solution that has a \(\mathrm{pH}\) of \(7.0 .\) The other half-cell is identical to the first except that the solution around the platinum electrode has a pH of \(0.0\). What is the cell voltage?

Which species in each pair is the stronger reducing agent? (a) \(\mathrm{Cr}\) or \(\mathrm{Cd}\) (b) \(\mathrm{I}^{-}\) or \(\mathrm{Br}^{-}\) (c) \(\mathrm{OH}^{-}\) or \(\mathrm{NO}_{2}^{-}\) (d) NO in acidic solution or NO in basic solution

Which of the following reactions is (are) spontaneous at standard conditions? (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+6 \mathrm{Cl}^{-}(a q) \longrightarrow\) \(2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{Cl}_{2}(g)\) (b) \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{Cl}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}_{2}(g)\) (c) \(3 \mathrm{Fe}(s)+2 \mathrm{AuCl}_{4}^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+8 \mathrm{Cl}^{-}(a q)+3 \mathrm{Fe}^{2+}(a q)\)

Consider a voltaic cell at \(25^{\circ} \mathrm{C}\) in which the following reaction takes place. $$ 3 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+6 \mathrm{H}^{+}(a q)+2 \mathrm{Au}(s) \longrightarrow 2 \mathrm{Au}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O} $$ (a) Calculate \(E\). (b) Write the Nernst equation for the cell. (c) Calculate \(E\) when \(\left[\mathrm{Au}^{3+}\right]=0.250 \mathrm{M},\left[\mathrm{H}^{+}\right]=1.25 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=\) \(1.50 M\)

An electrolytic cell consists of a \(100.0-\mathrm{g}\) strip of copper in \(0.200 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) and a \(100.0-\mathrm{g}\) strip of \(\mathrm{Cr}\) in \(0.200 \mathrm{M} \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\). The overall reac- tion is: $$ 3 \mathrm{Cu}(s)+2 \mathrm{Cr}^{3+}(a q) \longrightarrow 3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cr}(s) \quad E^{\circ}=-1.083 \mathrm{~V} $$ An external battery provides 3 amperes for 70 minutes and 20 seconds with \(100 \%\) efficiency. What is the mass of the copper strip after the battery has been disconnected?
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