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A "silver coulometer" (Study Question 72) was used in the past to measure the current flowing in an electrochemical cell. Suppose you found that the current flowing through an electrolysis cell deposited \(0.089 \mathrm{g}\) of \(\mathrm{Ag}\) metal at the cathode after exactly 10 min. If this same current then passed through a cell containing gold(III) ion in the form of \(\left(\mathrm{AuCl}_{4}\right)^{-}\), how much gold was deposited at the cathode in that electrolysis cell?

Short Answer

Expert verified
0.0541 g of gold was deposited.

Step by step solution

01

Calculate Moles of Silver Deposited

First, find the moles of silver deposited using its molar mass. The molar mass of silver (Ag) is approximately 107.87 g/mol. \[ \text{Moles of Ag} = \frac{0.089 \text{ g}}{107.87 \text{ g/mol}} \approx 0.000825 \text{ mol} \]
02

Determine Charge Passed Through the Cell

Use Faraday's law to determine the total charge. Silver has a valency of +1, so 1 mole of Ag requires 1 mole of electrons (or 1 Faraday, which is 96485 C) to be deposited.\[ Q = \text{Moles of Ag} \times 96485 \text{ C/mol} \approx 0.000825 \times 96485 \approx 79.58 \text{ C} \]
03

Current Through the Cell

The current can be calculated using the formula \( I = \frac{Q}{t} \) where \( t \) is the time in seconds (10 minutes = 600 seconds).\[ I = \frac{79.58 \text{ C}}{600 \text{ s}} \approx 0.1326 \text{ A} \]
04

Calculate Moles of Gold Deposited

Gold(III) ions \((\mathrm{AuCl}_{4})^{-}\) require 3 moles of electrons for each mole of gold deposited. Use the previously calculated charge to determine moles of gold.\[ \text{Moles of Au} = \frac{79.58 \text{ C}}{3 \times 96485 \text{ C/mol}} \approx 0.000275 \text{ mol} \]
05

Convert Moles of Gold to Grams

The molar mass of gold (Au) is approximately 196.97 g/mol. Convert the moles of gold to grams.\[ \text{Mass of Au} = 0.000275 \text{ mol} \times 196.97 \text{ g/mol} \approx 0.0541 \text{ g} \]

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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 a fundamental principle in electrochemistry that describes the relationship between electric charge and the material deposited during electrolysis. It states that the amount of substance formed at an electrode during electrolysis is directly proportional to the total electric charge passed through the cell.
This law can be expressed mathematically as \( Q = n \times F \), where \( Q \) is the charge in coulombs, \( n \) represents the moles of electrons involved, and \( F \) is Faraday's constant, approximately 96485 C/mol.
  • This relationship helps us calculate the amount of metal deposited in an electrolytic process by relating it to the charge that flows through the system.
  • For example, in the case of silver deposition, Faraday's Law allows us to determine the charge based on the known mass and molar mass of silver.
  • This understanding is crucial for predicting how much of a substance will accumulate during electrolysis.
Electrolysis
Electrolysis is a chemical process where electrical energy is used to drive a non-spontaneous reaction, typically resulting in the decomposition of compounds.

During electrolysis, an electric current is passed through an electrolyte, causing ions to move toward their respective electrodes and redox reactions to occur.
  • At the cathode, reduction takes place where cations gain electrons and get deposited as metal.
  • At the anode, oxidation may result, often involving the loss of electrons from anions.
This process underpins the functioning of electrochemical cells and is especially useful in refining and electroplating metals.

The exercise involving the deposition of silver and gold ions showcases these principles by illustrating the conversion of metallic ions to their solid metal forms at the cathode.
Coulometer
A coulometer is an instrument used to measure the electric charge (in coulombs) transferred in an electrochemical reaction. It's an essential tool for analyzing the time and current involved in electrolysis processes.

Coulometers help ensure accuracy by providing precise measurements of the total amount of charge that has passed through a system.
  • In the original exercise, the silver coulometer was used to verify the current in an electrochemical cell based on the mass of silver deposited.
  • This method relies on the predetermined relationship between charge and mass, where a specific amount of charge results in a predictable amount of material deposition.
  • This tool is critical for labs and industries where precise electrochemical calculations are necessary.
Valency
Valency refers to the combining power of an element, often indicated by the number of electrons an atom can lose, gain, or share. In electrochemistry, understanding valency is critical for determining how many electrons are involved in an electrochemical reaction.
For instance, silver (\(Ag\)) has a valency of +1, meaning each atom requires one electron for reduction. In contrast, gold ions (\((AuCl_4)^-\)) have a valency linked to gold being in the +3 oxidation state, as they require three electrons to be reduced to metallic gold.

When calculating deposition amounts during electrolysis, knowing valency helps determine the required charge and electrons per atom or molecule.
This is key in the original problem, where calculations hinge on the fact that gold ions require more electrons for reduction due to their higher valency relative to silver.
Current Calculation
Current (I), measured in amperes (A), is a crucial parameter in electrochemistry, representing the rate of charge flow. When conducting electrolysis, accurate current measurement is essential to determine the resultant chemical changes at electrodes.
The current can be calculated using the formula \( I = \frac{Q}{t} \), where \( Q \) is the charge in coulombs and \( t \) is time in seconds. In the given problem, the current was found by measuring the mass of deposited metal and calculating the charge involved.
  • A precise understanding of current is essential for predicting the deposition rate and understanding the overall energy consumption of the process.
  • In electrochemical cells, maintaining a steady current ensures a consistent output of the deposition process.
By mastering current calculations, one can effectively control and predict outcomes in various electrochemical reactions.

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Most popular questions from this chapter

A Balance the following equations involving organic compounds. (a) \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Ag}(\mathrm{s})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \quad\) (acid solution) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq}) \quad\) (acid solution)

A Write balanced equations for the following reduction half-reactions involving organic compounds. (a) \(\mathrm{HCO}_{2} \mathrm{H} \longrightarrow \mathrm{CH}_{2} \mathrm{O} \quad\) (acid solution) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} \quad\) (acid solution) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \quad\) (acid solution) (d) \(\mathrm{CH}_{3} \mathrm{OH} \longrightarrow \mathrm{CH}_{4} \quad\) (acid solution)

The half-cells \(\mathrm{Ag}^{+}(\mathrm{aq}) | \mathrm{Ag}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g}) | \mathrm{Cl}^{-}(\mathrm{aq})\) are linked to create a voltaic cell. (a) Write equations for the oxidation and reduction halfreactions and for the overall (cell) reaction. (b) Which half-reaction occurs in the anode compartment and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the ___ electrode to the ___ electrode. Negative ions move in the salt bridge from the ___ half-cell to the ___ half-cell.

One half-cell in a voltaic cell is constructed from a copper wire dipped into a \(4.8 \times 10^{-3} \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) The other half-cell consists of a zinc electrode in a \(0.40 \mathrm{M}\) solution of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the cell potential.

Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{CrO}_{4}^{2-}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s}) \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) (d) \(\mathrm{HS}^{-}(\mathrm{aq})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq})\)

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