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Chapter 21: Problem 114

Brass, an alloy of copper and zinc, can be produced by simultaneously electroplating the two metals from a solution containing their \(2+\) ions. If \(65.0 \%\) of the total current is used to plate copper, while \(35.0 \%\) goes to plating zinc, what is the mass percent of copper in the brass?

Short Answer

Expert verified
64.27%

Step by step solution

01

Identify the given information

From the problem, it is provided that 65.0% of the current is used to plate copper and 35.0% to plate zinc.
02

Understand the relation of current usage

The current usage directly correlates with the amount of each metal being plated since the amount of electrolysis product is proportional to the electric charge passed.
03

Use Faraday's Law of Electrolysis

Faraday's Law states that the mass of substance deposited at an electrode is directly proportional to the charge: \[ m = \frac{Q \times M}{n \times F} \]where \( Q \) is the total charge, \( M \) is the molar mass, \( n \) is the number of electrons in the reduction, and \( F \) is Faraday's constant (approximately 96485 C/mol).
04

Separate the charge going to copper and zinc

Assume a total charge of \( Q \). Therefore, the charge used for plating copper is \( 0.65Q \) and for zinc is \( 0.35Q \).
05

Compute the masses of copper and zinc plated

For copper (Cu, with molar mass M = 63.5 g/mol and n = 2):\[ m_{Cu} = \frac{0.65Q \times 63.5}{2 \times 96485} \]For zinc (Zn, with molar mass M = 65.5 g/mol and n = 2):\[ m_{Zn} = \frac{0.35Q \times 65.5}{2 \times 96485} \]
06

Calculate the ratio of the masses

Divide each expression by \( Q \):For copper,\[ m_{Cu} = \frac{0.65 \times 63.5}{2 \times 96485}Q \]For zinc,\[ m_{Zn} = \frac{0.35 \times 65.5}{2 \times 96485}Q \]
07

Find mass of plated metals

Since the ratio is applicable to any \( Q \), normalize without it:\[ m_{Cu} \times 2 \times 96485 = 0.65 \times 63.5 \];\[ m_{Zn} \times 2 \times 96485 = 0.35 \times 65.5 \];\[ m_{Cu} = 0.0002143Q \]; \[ m_{Zn} = 0.00011906Q \].
08

Compute the mass percent of copper

Given the masses of copper and zinc, sum their weights:\[ (0.0002143 + 0.00011906)Q = 0.00033336 Q \]. So the mass percent of Cu in brass is\[ \frac{m_{Cu}}{m_{Cu}+m_{Zn}} \times 100 \ =~ 64.27%\].

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Key Concepts

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

Faraday's Law
Faraday's Law is an essential concept for understanding electroplating. It states that the mass of a substance deposited at an electrode is directly proportional to the electric charge passed through the substance. The equation for Faraday's Law is: \[ m = \frac{Q \times M}{n \times F} \]. Here, \( m \) is the deposited mass, \( Q \) is the total charge, \( M \) is the molar mass of the substance, \( n \) is the number of electrons involved in the reduction or oxidation reaction, and \( F \) is Faraday's constant, approximately 96485 C/mol. This law helps us understand how the current used in electrolysis corresponds to the mass of metal deposited on the electrode. Knowing this, students can solve problems related to electroplating by calculating the amount of metal deposited using given current and molar masses.
Mass Percent Calculation
Mass percent is an important concept in chemistry, particularly when dealing with alloys like brass. Mass percent represents the concentration of an element in a compound or mixture. It is calculated using the formula: \[ \text{Mass \, Percent} = \left( \frac{\text{Mass of Component}}{\text{Total Mass}} \right) \times 100 \% \]. In the provided exercise, the mass percent of copper in brass is determined by calculating the masses of copper and zinc deposited through electroplating and then finding their respective percentages of the total mass. The completed calculation in the exercise shows that if 65% of the current is used to deposit copper and 35% to deposit zinc, the mass percent of copper in brass is found to be approximately 64.27%. This is done by first calculating the masses of copper and zinc and then using their sum as the total mass for the mass percent formula.
Alloy Production
Alloys are mixtures of two or more elements, where at least one element is a metal. Brass is an example of an alloy, composed mainly of copper and zinc. Producing alloys often involves processes like electroplating, in which metals are deposited onto a surface using electrical current. When electroplating copper and zinc to produce brass, the proportion of each metal plated depends on the current distribution. By knowing the percentage of current used for each metal, the mass of each metal in the final alloy can be calculated, leading to the determination of the alloy's composition. This method ensures a controlled ratio of metals, resulting in desired physical properties for the alloy. Understanding this process is crucial for industries where precise alloy composition is necessary for specific applications, such as in manufacturing, electrical engineering, and metallurgy.

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Most popular questions from this chapter

Comparing the standard electrode potentials \(\left(E^{\circ}\right)\) of the Group \(1 \mathrm{~A}(1)\) metals \(\mathrm{Li}, \mathrm{Na},\) and \(\mathrm{K}\) with the negative of their first ionization energies reveals a discrepancy: Ionization process reversed: \(\mathrm{M}^{+}(g)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(g)\) Electrode reaction: $$\mathrm{M}^{+}(a q)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(s)$$ $$\begin{array}{lcc}\text { Metal } & -\text { IE (kJ/mol) } & E^{\circ} \text { (V) } \\\\\hline \text { Li } & -520 & -3.05 \\\\\text { Na } & -496 & -2.71 \\\ \text { K } & -419 & -2.93\end{array}$$ Note that the electrode potentials do not decrease smoothly down the group, whereas the ionization energies do. You might expect that, if it is more difficult to remove an electron from an atom to form a gaseous ion (larger IE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smaller \(E^{\circ}\) ), yet \(\mathrm{Li}^{+}(a q)\) is more difficult to reduce than \(\mathrm{Na}^{+}(a q) .\) Applying Hess's law, use an approach similar to a Born-Haber cycle to break down the process occurring at the electrode into three steps, and label the energy involved in each step. How can you account for the discrepancy?

Balance the following skeleton reactions, and identify the oxidizing and reducing agents: (a) \(\mathrm{O}_{2}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{3}^{-}(a q)[\) acidic \(]\) (b) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)+\mathrm{Cu}(\mathrm{OH})_{2}(s)[\) basic \(]\) (c) \(\mathrm{AsO}_{4}^{3-}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{AsO}_{2}^{-}(a q)+\mathrm{NO}_{3}^{-}(a q)[\) basic \(]\)

Aqua regia, a mixture of concentrated \(\mathrm{HNO}_{3}\) and \(\mathrm{HCl}\) was developed by alchemists as a means to "dissolve" gold. The process is a redox reaction with this simplified skeleton reaction: $$\mathrm{Au}(s)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AuCl}_{4}^{-}(a q)+\mathrm{NO}_{2}(g)$$ (a) Balance the reaction by the half-reaction method. (b) What are the oxidizing and reducing agents? (c) What is the function of \(\mathrm{HCl}\) in aqua regia?

What are \(E_{\text {cell }}^{\circ}\) and \(\Delta G^{\circ}\) of a redox reaction at \(25^{\circ} \mathrm{C}\) for which \(n=2\) and \(K=65 ?\)

Electrolysis of molten \(\mathrm{MgCl}_{2}\) is the final production step in the isolation of magnesium from seawater by the Dow process (Section 22.4). Assuming that \(45.6 \mathrm{~g}\) of Mg metal forms, (a) How many moles of electrons are required? (b) How many coulombs are required? (c) How many amps will produce this amount in \(3.50 \mathrm{~h} ?\)
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