/*! 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 108 A spoon was silver-plated electr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A spoon was silver-plated electrolytically in an \(\mathrm{AgNO}_{3}\) solution. (a) Sketch a diagram for the process. (b) If \(0.884 \mathrm{~g}\) of Ag was deposited on the spoon at a constant current of \(18.5 \mathrm{~mA}\), how long (in min) did the electrolysis take?

Short Answer

Expert verified
The electrolysis took about 713 minutes.

Step by step solution

01

Understanding Electroplating

Electroplating involves coating an object with a metal using electrical current. The object to be coated is the cathode while the metal that coats it comes from a metal salt solution; here, it is an AgNO extsubscript{3} solution. During the process, Ag extsuperscript{+} ions in solution gain electrons and form metallic silver on the cathode surface.
02

Sketch a Diagram for Electroplating

Draw a cell with two electrodes submerged in AgNO extsubscript{3} solution. Connect the electrodes to an external power source. Label one electrode as the "Anode" (Ag metal piece) and the other as the "Cathode" (spoon). Ag atoms from the anode dissolve into the solution, and Ag extsuperscript{+} ions plate onto the cathode (spoon).
03

Apply Faraday's Laws of Electrolysis

To find the time for electrolysis, use the formula: \( t = \frac{m \times F}{I \times M} \), where \(t\) is time, \(m\) is mass of Ag deposited (0.884 g), \(F\) is Faraday's constant (96485 C/mol), \(I\) is current (18.5 mA or 0.0185 A), and \(M\) is molar mass of Ag (107.87 g/mol).
04

Calculate the Number of Moles of Ag

Determine the moles of Ag deposited using its mass and molar mass: \( \text{moles of Ag} = \frac{0.884 \text{ g}}{107.87 \text{ g/mol}} = 0.00819 \text{ mol} \).
05

Calculate the Charge Required

Calculate total charge required for the deposition of Ag using \( ext{Q} = n imes F \), where \(n\) is moles of Ag (0.00819 mol) and \(F\) is Faraday's constant (96485 C/mol): \( Q = 0.00819 \text{ mol} \times 96485 \text{ C/mol} = 790.6 \text{ C} \).
06

Calculate the Time of Electrolysis

Use the total charge and current to find the time: \( t = \frac{Q}{I} = \frac{790.6 \text{ C}}{0.0185 \text{ A}} = 42789 \text{ s} \). Convert this to minutes by dividing by 60: \( t = \frac{42789}{60} \approx 713.15 \text{ minutes} \).
07

Final Result

The electrolysis took approximately 713 minutes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Faraday's Laws of Electrolysis
Faraday's Laws of Electrolysis are key principles in understanding how electroplating works. These laws explain the relationship between the amount of electric charge used in the process and the amount of substance that is deposited.
Faraday's First Law states that the mass of a substance deposited at an electrode is directly proportional to the amount of electric charge passed through the electrolyte. This means more current or longer time results in more deposition.
Faraday's Second Law states that for the same quantity of electricity, the masses of substances deposited are directly proportional to their equivalent weights. Equivalent weight is obtained by dividing the molar mass by the number of electrons exchanged in the reaction.
In the case of silver plating, Faraday's laws help us calculate how much metallic silver will deposit based on the charge transferred, which is crucial for predicting how long the process takes.
Silver plating
Silver plating is the process of depositing a thin layer of silver onto an object, often to improve its appearance or corrosion resistance. This involves immersing the object, called the cathode, into a solution containing silver ions, such as $ ext{AgNO}_3$ (Silver Nitrate).
In this setup:
  • The object to be plated (such as a spoon) serves as the cathode.
  • The silver metal, usually in bar form, acts as the anode.
  • The electrolyte is a silver salt solution, such as $ ext{AgNO}_3$.
The silver ions ($ ext{Ag}^+$) in the solution are attracted to the cathode. As the electrical current flows, these ions accept electrons and form a thin layer of metallic silver on the object.
This process improves the object's aesthetic value and durability by providing a silver finish that is both attractive and protective.
Electrolysis calculation
Electrolysis calculations involve determining various quantities related to the deposition process, such as the time required, the mass of metal deposited, and the charge needed.
To solve an electrolysis problem, one typically follows these steps:
  • Calculate the moles of metal deposited using its mass and molar mass: ext{moles} = rac{ ext{mass}}{ ext{molar mass}}.
  • Determine the charge required using ext{Q} = ext{moles} imes ext{F}, where ext{F} is Faraday's constant (96485 C/mol).
  • Find the time required by dividing the total charge by the current: ext{time} = rac{ ext{Q}}{ ext{I}}, converting the result to the desired time units.
For example, if 0.884 g of silver is plated with a constant current of 18.5 mA, these steps will determine how long the process will take.
Calculations based on these steps ensure that the desired thickness of the metal coating can be achieved reliably.
Electrochemical cell
An electrochemical cell is a device capable of generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy.
In silver plating, the main component is an electrolytic cell, which includes:
  • An anode, made of silver metal, where oxidation takes place, releasing $ ext{Ag}^+$ ions into the solution.
  • A cathode, such as the spoon, where the $ ext{Ag}^+$ ions are reduced, forming metallic silver coatings.
  • An electrolyte, like $ ext{AgNO}_3$, which contains the ions needed for the reaction to proceed.
The external power supply is crucial as it drives the movement of electrons and facilitates the deposition of silver on the cathode.
This setup illustrates the fundamental principles of electrochemical cells and showcases their practical application in processes such as electroplating.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

"Galvanized iron" is steel sheet that has been coated with zinc; "tin" cans are made of steel sheet coated with tin. Discuss the functions of these coatings and the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface of a galvanized iron sheet or a tin can.

A steady current was passed through molten \(\operatorname{CoSO}_{4}\) until \(2.35 \mathrm{~g}\) of metallic cobalt was produced. Calculate the number of coulombs of electricity used.

The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable: The net transformation is \(\mathrm{Zn}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnO}(s)\) (a) Write the half-reactions at the zinc-air electrodes, and calculate the standard emf of the battery at \(25^{\circ} \mathrm{C}\). (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from \(1 \mathrm{~kg}\) of the metal) of the zinc electrode? (d) If a current of \(2.1 \times 10^{5} \mathrm{~A}\) is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the battery every second? Assume that the temperature is \(25^{\circ} \mathrm{C}\) and the partial pressure of oxygen is 0.21 atm.

Given that \(E^{\circ}=0.52 \mathrm{~V}\) for the reduction \(\mathrm{Cu}^{+}(a q)+e^{-}\) \(\longrightarrow \mathrm{Cu}(s),\) calculate \(E^{\circ}, \Delta G^{\circ},\) and \(K\) for the following reaction at \(25^{\circ} \mathrm{C}\) : $$ 2 \mathrm{Cu}^{+}(a q) \rightleftarrows \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s) $$

Calculate the amounts of \(\mathrm{Cu}\) and \(\mathrm{Br}_{2}\) produced in \(1.0 \mathrm{~h}\) at inert electrodes in a solution of \(\mathrm{CuBr}_{2}\) by a current of \(4.50 \mathrm{~A}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.