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Chapter 25: Problem 62

Suppose that it is planned to electrodeposit \(200.0 \mathrm{mg}\) of gold onto the surface of a steel object via the process $$ \left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Au}(s)+2 \mathrm{CN}^{-}(a q) $$ If the electric current in the circuit is set at \(30.0 \mathrm{~mA}\) for how long should the current be passed?

Short Answer

Expert verified
The current should be passed for approximately 3263 seconds.

Step by step solution

01

Calculate the moles of gold

To find out how many moles of gold we have, we need to convert the mass of gold from milligrams to grams and then use the molar mass of gold. First, convert the mass: \[ 200.0 \text{ mg} = 0.200 \text{ g} \]Now, use the molar mass of gold (\(196.97 \text{ g/mol} \)) to find the moles of gold:\[ \text{moles of Au} = \frac{0.200 \text{ g}}{196.97 \text{ g/mol}} \approx 0.001015 \text{ mol} \]
02

Use Faraday's Law of Electrolysis

From Faraday's Law, the amount of substance liberated at an electrode is directly proportional to the number of moles of electrons transferred. As per the given equation, 1 mole of \( \left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-} \) requires 1 mole of electrons to deposit 1 mole of \( \mathrm{Au} \).Thus, for \( 0.001015 \text{ mol} \) of \( \mathrm{Au} \), the same amount \( 0.001015 \text{ mol} \) of electrons \(\text{e}^{-}\) is required.
03

Calculate the total charge needed

Each mole of electrons carries a charge of \(96485 \text{ C/mol}\) (Faraday's constant).Calculate the total charge \(Q\) required for the deposition:\[ Q = \text{moles of electrons} \times 96485 \text{ C/mol} \]\[ Q = 0.001015 \text{ mol} \times 96485 \text{ C/mol} \approx 97.9 \text{ C} \]
04

Determine the time for current to flow

The total time \( t \) for which the current must flow can be calculated using the formula:\[ t = \frac{Q}{I} \]where \( I = 30.0 \text{ mA} = 0.0300 \text{ A} \).Thus,\[ t = \frac{97.9 \text{ C}}{0.0300 \text{ A}} \approx 3263 \text{ seconds} \]

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Key Concepts

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

Faraday's Law of Electrolysis
Faraday's Law of Electrolysis is a fundamental concept in electrochemistry. It describes how electric charge relates to the amount of substance deposited or dissolved at an electrode during electrolysis. According to this law: - The amount of chemical change is proportional to the quantity of electricity that passes through the circuit. - One mole of electrons transfers a charge of approximately 96485 coulombs, known as Faraday's constant. This means, if you know the amount of charge required, you can calculate how much of a substance will deposit or dissolve.

Faraday's law is particularly powerful because it directly connects the atomic scale (moles of atoms) with the macroscopic world (the current in a wire). In our exercise, this law helps us determine how long a current should flow to deposit a specific mass of gold.

When tackling an electrodeposition problem, identifying the required charge using Faraday's Law allows you to link electric parameters with material outcomes.
Mole Calculations
Mole calculations are essential in understanding how much of a substance is involved in a chemical reaction, such as during electrolysis. A mole is a unit used to quantify the amount of a chemical substance.- To find the number of moles, use the formula: \[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]- The molar mass is the mass of one mole of a substance, given in grams per mole.In the exercise, we computed the moles of gold by first converting its mass from milligrams to grams, then dividing by the molar mass of gold, which is 196.97 g/mol. This provided us with the moles of gold needed for electrodeposition.

Being thorough with mole calculations is crucial as they lay the groundwork for further calculations involving chemical reactions, such as determining how much charge is needed for electrolysis.
Gold Electrodeposition
Gold electrodeposition is a process used to coat a surface with a thin layer of gold using electrical energy. During this process, gold ions in an electrolyte solution gain electrons and are reduced to form solid gold on an object.

The chemical reaction can be represented as follows:\[ \left[\mathrm{Au} (\mathrm{CN})_{2} \right]^{-} + \mathrm{e}^{-} \rightarrow \mathrm{Au} + 2 \mathrm{CN}^{-} \]- Gold electrodeposition is often used for decorative purposes and to improve corrosion resistance.- It's essential to calculate the right amount of electricity (using Faraday's Law) to achieve the desired gold coating.In our exercise, we determined the mass of gold to be deposited and calculated how long the current should flow to achieve this deposition using the information about current and charge needed.

Understanding the intricacies of gold electrodeposition not only requires knowledge of electrochemical principles but also the practical application of those principles in industrial processes.

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Most popular questions from this chapter

Calculate the voltage at \(25^{\circ} \mathrm{C}\) of an electrochemical cell for the reaction described by the equation $$ \begin{aligned} \mathrm{Cd}(s)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q,&0.10 \mathrm{M}) \leftrightharpoons \\ & \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(a q, 0.010 \mathrm{M})+\mathrm{Pb}(s) \end{aligned} $$ See Appendix \(\mathrm{G}\) for the necessary \(E^{\circ}\) data.

The voltage generated by the zinc concentration cell described by $$ \mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}(a q, 0.100 \mathrm{M}) \| \mathrm{Zn}^{2+}(a q)\right| \mathrm{Zn}(s) $$ is \(20.0 \mathrm{mV}\) at \(25^{\circ} \mathrm{C}\). Calculate the concentration of the \(\mathrm{Zn}^{2+}(a q)\) ion at the cathode.

The cell diagram for an electrochemical cell is given as $$ \operatorname{In}(s)\left|\operatorname{In}\left(\mathrm{ClO}_{4}\right)_{3}(a q) \| \mathrm{CdCl}_{2}(a q)\right| \operatorname{Cd}(s) $$ Write the equations for the half reactions that occur at the two electrodes and the net cell reaction.

The standard voltage of the cell described by $$ \operatorname{Pt}(s)\left|\mathrm{H}_{2}(g)\right| \mathrm{H}^{+}(a q) \| \mathrm{Cd}^{2 *}(a q) \mid \mathrm{Cd}(s) $$ is \(-0.403 \mathrm{~V}\). Write the cell equation and calculate the cell voltage when \(\left[\mathrm{H}^{+}\right]=0.10 \mathrm{M}, P_{\mathrm{H}_{2}}=0.10 \mathrm{bar}\), and \(\left[\mathrm{Cd}^{2+}\right]=2.5 \times 10^{-3} \mathrm{M}\) at \(25^{\circ} \mathrm{C}\)

Given the standard reduction voltages at \(25^{\circ} \mathrm{C}\) for the half- reaction equations $$ \begin{array}{ll} \mathrm{Cr}^{3+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Cr}^{2-}(a q) & E_{\mathrm{red}}^{\circ}=-0.407 \mathrm{~V} \\ \mathrm{Cr}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(s) & E_{\mathrm{red}}^{\circ}=-0.913 \mathrm{~V} \end{array} $$ use these data to determine the standard reduction voltage at \(25^{\circ} \mathrm{C}\) for the half-reaction equation $$ \mathrm{Cr}^{\mathrm{s}+}(a q)+\mathrm{3e}^{-} \rightarrow \mathrm{Cr}(s) $$ Compare your answer to the value listed in Appendix G. Why can't we simply add the reduction voltages of the two half-reaction equations above to get this value?
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