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Chapter 1: Problem 21

A chemist wants to plate out 1.00 g of solid lead from a solution containing aqueous \(\mathrm{Pb}^{2+}\) ions. Which of the following expressions will equal the amount of time, in seconds, it takes if a current of 5.00 \(\mathrm{A}\) is applied? (A) \(\frac{(2)(55.85)(5.00)}{96500}\) (B) \(\frac{(2)(96500)}{(55.85)(5.00)}\) (C) \(\frac{(55.85)(96500)}{(2)(5.00)}\) (D) \(\frac{(2)(55.85)(96500)}{(5.00)}\)

Short Answer

Expert verified
The correct expression that equals the time in seconds is (D) \(\frac{(2)(55.85)(96500)}{(5.00)}\).

Step by step solution

01

Understanding Faraday's law

Faraday's Law states that when one mole of a substance is deposited at an electrode by electrolysis, the quantity of electricity involved is one Faraday (F) which equals 96500 coulombs. The formula relating these quantities is \[Q = n * F\], where Q is the charge in coulombs, n is the amount in moles and F is the Faraday constant. Since \(Q = I * t\), where I is the current in amperes and t is the time in seconds, we can substitute Q in the equation to find the time t as \[t = n * F / I\]
02

Calculating the amount of lead in moles

First, we need to convert the mass of lead to moles. Lead ions (Pb^2+) will require 2 electrons to be reduced to solid lead. The atomic mass of lead is 207.2 g/mol, so 1 g of lead is \(\frac{1.00}{207.2}\) = 0.00483 moles. The number of moles (n) becomes 2*0.00483 = 0.00966 moles.
03

Substitute values into the time formula and compute

Next, we substitute n = 0.00966 moles, F = 96500 C/mol and I = 5.00 A into the formula for time \[t = n * F / I\]. This gives the time as \[ t = (0.00966 * 96500) / 5.00 = 186.3251 seconds\]
04

Determine the matching expression

Finally, we look through the given options to determine which expression equals to the time in seconds that we calculated. Given the values we used: 2 for the electrons, 0.00483 for the moles of lead, 96500 for the Faraday constant and 5.00 for the current, it is option D, \(\frac{(2)(0.00483)(96500)}{(5.00)}\) that equals our calculated time.

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Most popular questions from this chapter

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause a reduction in the value for the equilibrium constant? (A) Increasing the amount of \(\mathrm{SO}_{3}\) (B) Reducing the amount of \(\mathrm{O}_{2}\) (C) Raising the temperature (D) Lowering the temperature

In which of the following reactions is entropy increasing? (A) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) (B) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) (C) \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{HCl}(g)\) (D) \(2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\)

If the solubility of \(\mathrm{BaF}_{2}\), is equal to \(x,\) which of the following expressions is equal to the solubility product, \(K_{s p}\) , for \(\operatorname{BaF}_{2} ?\) (A) \(x^{2}\) (B) 2\(x^{2}\) (C) 2\(x^{3}\) (D) 4\(x^{3}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)
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