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Chapter 18: Problem 103

It took \(2.30\) min using a current of \(2.00 \mathrm{~A}\) to plate out all the silver from \(0.250 \mathrm{~L}\) of a solution containing \(\mathrm{Ag}^{+}\). What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?

Short Answer

Expert verified
The original concentration of \(Ag^+\) ions in the solution was \(0.0114\,\mathrm{M}\).

Step by step solution

01

Determine the charge transferred using the current and time

Given that the current is \(I = 2.00\,\mathrm{A}\), and the time duration is \(t = 2.30\,\mathrm{min}\), we can find the total charge (Q) transferred during the plating process. Convert the time to seconds and use the formula \(Q = It\), which gives: \(Q = (2.00\,\mathrm{A})(2.30\,\mathrm{min} \times 60\,\mathrm{s/min})\) \(Q = 276\,\mathrm{C}\)
02

Calculate the moles of silver ions plated out

Use Faraday's law, which relates the quantity of a substance undergoing electrolysis to the charge transferred: \(n = \frac{Q}{zF}\) Here, n represents the moles of silver ions plated out, z is the charge number (+1 for \(Ag^+\) ions), and F is the Faraday constant (\(96485\,\mathrm{C/mol}\)). Calculate the moles of silver ions: \(n=\frac{276\,\mathrm{C}}{1\times96485\,\mathrm{C/mol}}\) \(n = 0.00286\,\mathrm{mol}\)
03

Find the original concentration of Ag(+) in the solution

We are given the volume of the solution \(V = 0.250\,\mathrm{L}\), and we have calculated the moles of \(Ag^+\) ions plated out (n). Divide the moles (n) by the volume (V) to find the initial concentration of \(Ag^+\) ions in the solution: \(c_{Ag^+} = \frac{n}{V}\) \(c_{Ag^+} = \frac{0.00286\,\mathrm{mol}}{0.250\,\mathrm{L}}\) \(c_{Ag^+} = 0.0114\,\mathrm{M}\) So, the original concentration of \(Ag^+\) ions in the solution was \(0.0114\,\mathrm{M}\).

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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 central to the process of electrolysis, as it relates the amount of substance transformed at an electrode to the amount of electric charge passed through the electrolyte. The essence of Faraday's Law is captured in the formula:
  • \( n = \frac{Q}{zF} \)
where \( n \) is the moles of substance, \( Q \) represents the total charge, \( z \) is the charge number (specific to the ions involved), and \( F \) is Faraday's constant, approximately \( 96485 \mathrm{~C/mol} \).

In practical terms, this means that the number of moles of a substance appearing at the electrode is proportional to the charge passing through the electrolyte. This relationship helps us determine the amount of material altered in processes like electroplating or electrolysis, where ions deposit as atoms at the electrodes.
Electrolysis
Electrolysis is a chemical process that uses electric current to drive a non-spontaneous chemical reaction. Unlike spontaneous reactions, which occur naturally, electrolysis requires the input of energy from an external source, such as a battery.

In the context of our exercise, electrolysis was used to plate silver out from a solution. An electric current of \(2.00 \mathrm{~A}\) was applied for \(2.30\) minutes, causing \(\mathrm{Ag}^{+}\) ions from the solution to be reduced and deposited as solid silver on the cathode.
  • Cathode: Where reduction takes place and positive ions gain electrons.
  • Anode: Where oxidation takes place and negative ions release electrons.
This specific process of plating or depositing a metal through the use of an electric current is known as electroplating. It is widely used in industries for numerous applications, including coating metals to prevent corrosion, enhancing appearance, and providing surface protection.
Concentration Calculation
Concentration calculations are essential in determining the quantity of solute in a solution. In our problem, determining the initial concentration of silver ions involved calculating the total moles of silver plated over a certain volume of solution. Once the moles (\(n\)) were determined using Faraday's Law, the concentration (\(c\)) was found by dividing the moles by the volume (\(V\)) of the solution:
  • \( c_{Ag^+} = \frac{n}{V} \)
  • \( c_{Ag^+} = \frac{0.00286 \mathrm{~mol}}{0.250 \mathrm{~L}} = 0.0114 \mathrm{~M} \)


Concentration is typically expressed in molarity (M), which is moles of solute per liter of solution. Understanding how to conduct these calculations is crucial in fields such as chemistry and biochemistry, where solution properties can significantly affect reactions and processes.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ In this cell, the silver compartment contains a silver electrode and excess \(\operatorname{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)\), and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 M\). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. Assuming \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons & \\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned} $$

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\).

Consider a concentration cell similar to the one shown in Exercise 69, except that both electrodes are made of \(\mathrm{Ni}\) and in the left-hand compartment \(\left[\mathrm{Ni}^{2+}\right]=1.0 M .\) Calculate the cell potential at \(25^{\circ} \mathrm{C}\) when the concentration of \(\mathrm{Ni}^{2+}\) in the compartment on the right has each of the following values. a. \(1.0 M\) b. \(2.0 \mathrm{M}\) c. \(0.10 \mathrm{M}\) d. \(4.0 \times 10^{-5} M\) e. Calculate the potential when both solutions are \(2.5 M\) in \(\mathrm{Ni}^{2+}\). For each case, also identify the cathode, anode, and the direction in which electrons flow.

What reactions take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{NiBr}_{2}\) b. molten \(\mathrm{AlF}_{3}\) c. molten \(\mathrm{MnI}_{2}\)

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ} &=0.34 \mathrm{~V} \\ \mathrm{~V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ} &=-1.20 \mathrm{~V} \end{aligned} $$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M\), and the vanadium compartment contains a vanadium electrode and \(\mathrm{V}^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{~L}\) of solution) was titrated with \(0.0800 M \mathrm{H}_{2} \mathrm{EDTA}^{2-}\), resulting in the reaction $$ \begin{array}{r} \mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \\ K=? \end{array} $$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{~mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{~V}\). The solution was buffered at a pH of \(10.00\). a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K\), for the titration reaction. c. Calculate \(\mathscr{B}_{\text {cell }}\) at the halfway point in the titration.
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