/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 The electrolysis of \(\mathrm{Bi... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 94

The electrolysis of \(\mathrm{BiO}^{+}\) produces pure bismuth. How long would it take to produce \(10.0 \mathrm{g}\) Bi by the electrolysis of a \(\mathrm{BiO}^{+}\) solution using a current of \(25.0 \mathrm{A} ?\)

Short Answer

Expert verified
It would take approximately \(184.46 \mathrm{s}\) to produce \(10.0 \mathrm{g}\) of bismuth by the electrolysis of a \(\mathrm{BiO}^{+}\) solution using a current of \(25.0 \mathrm{A}\).

Step by step solution

01

Calculate the moles of bismuth to be produced

First, we need to find the moles of bismuth (\(n_{Bi}\)) to be produced. To do this, divide the mass of bismuth (\(m_{Bi} = 10 g\)) by its molar mass (\(M_{Bi} = 208.98 g/mol\)): \[n_{Bi} = \frac{m_{Bi}}{M_{Bi}}\]
02

Determine the moles of electrons used in the process

In the electrolysis process, BiO鈦� ion loses one electron to form pure bismuth (Bi). Therefore, one mole of bismuth requires one mole of electrons. So, the moles of electrons needed (\(n_{e}\)) are equal to the moles of bismuth: \[n_{e} = n_{Bi}\]
03

Calculate the charge needed for the process

Now, we need to find the total charge (in coulombs) required for the electrolysis process (\(Q\)). Faraday's constant (\(F\)) states that one mole of electrons carries a charge of 96,485 C/mol. Therefore, to find the charge needed, multiply the moles of electrons by Faraday's constant: \[Q = n_{e} \times F\]
04

Calculate the time required for electrolysis

Finally, to find the time required for electrolysis (\(t\)), we will use the formula for electric current (\(I\)), which is the charge per unit time: \[I = \frac{Q}{t}\] Rearranging the equation to find the time, we get: \[t = \frac{Q}{I}\] Now, let's find the time by plugging in the given values: 1. Calculate the moles of bismuth to be produced: \[n_{Bi} = \frac{10.0 \mathrm{g}}{208.98 \mathrm{g/mol}} \approx 0.0478 \mathrm{mol}\] 2. Determine the moles of electrons used in the process: \[n_{e} = 0.0478 \mathrm{mol}\] 3. Calculate the charge needed for the process: \[Q = 0.0478 \mathrm{mol} \times 96485 \mathrm{C/mol} \approx 4611.47 \mathrm{C}\] 4. Calculate the time required for electrolysis: \[t = \frac{4611.47 \mathrm{C}}{25.0 \mathrm{A}} \approx 184.46 \mathrm{s}\] So, it would take approximately 184.46 seconds to produce 10.0 grams of bismuth by the electrolysis of a BiO鈦� solution using a current of 25.0 A.

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Key Concepts

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

Faraday's constant
Faraday's constant is an essential concept in electrochemistry, representing the charge of one mole of electrons. It is a derived value that makes calculations involving electrons more accessible. Faraday's constant is approximately 96,485 coulombs per mole (C/mol). This means that
  • 1 mole of electrons carries a charge of 96,485 C.
  • This value is crucial for converting between moles of electrons and electrical charge in an electrochemical cell.
If you're wondering why Faraday's constant is so vital, consider it as a bridge connecting the macroscopic world (moles of a substance) with the microscopic behavior of electrons in a chemical reaction. This constant ensures that when you calculate reactions or processes involving electrons, like electrolysis, you can smoothly translate those into measurable quantities of electric charge. In the context of our exercise, it is used to determine the total charge needed to produce a specified amount of bismuth by multiplying by the moles of electrons involved in the reaction.
Moles calculation
Calculating moles is fundamental in chemistry, as it allows you to quantify the number of particles, such as atoms or molecules, participating in a chemical reaction. Moles give chemists a consistent means of measuring amount of substance. The formula to calculate moles is:
  • Moles = Mass / Molar Mass
In our example:
  • The mass of bismuth, Bi, intended for production is 10 g.
  • The molar mass of bismuth is 208.98 g/mol, meaning each mole weighs 208.98 g.
By applying the moles formula, you can understand that:
  • Moles of bismuth (\(n_{Bi}\)) produced is \(\frac{10 \text{ g}}{208.98 \text{ g/mol}} \approx 0.0478 \text{ mol}\).
Calculating moles this way not only helps you determine how much bismuth can be produced, but it also allows further calculations like determining the moles of electrons needed for the electrolysis process.
Electric current calculation
Electric current is a measure of the flow of electric charge through a conductor, and it plays a crucial role in electrolysis. It is defined by the formula:
  • Current (I) = Charge (Q) / Time (t)
Where:
  • I is the current in amperes (A),
  • Q is the charge in coulombs (C),
  • t is the time in seconds (s).
This relationship shows that the amount of charge moved through the system over time determines the current. In electrolysis:
  • For our task, we need to calculate how long it takes to produce 10 g of bismuth using a current of 25 A.
  • Given the total charge needed is 4611.47 C, the formula rearranges to find time: \( t = \frac{Q}{I} \).
By inserting the known values, \( t = \frac{4611.47 C}{25.0 A} \approx 184.46 \text{ seconds} \).Understanding this equation helps to grasp the relationship between charge, time, and current in the process of electrolysis, ensuring that calculations align with real-world measurements.

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Most popular questions from this chapter

Consider the cell described below: $$ \mathrm{Al}\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Al}^{3+}\right]\) to have changed by \(0.60 \mathrm{mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]\) \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} &\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10 \(M\) NaOH that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\mathrm{For} \mathrm{Cu}(\mathrm{OH})_{2}\right.\) \(K_{\mathrm{sp}}=1.6 \times 10^{-19} \cdot \mathrm{J}\)

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{kJ} .\) What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.
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