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Chapter 20: Problem 118

An aqueous solution of an unknown salt of vanadium is electrolyzed by a current of \(2.50\) amps for \(1.90\) hours. The electroplating is carried out with an efficiency of \(95.0 \%\), resulting in a deposit of \(2.850 \mathrm{~g}\) of vanadium. a. How many faradays are required to deposit the vanadium? b. What is the charge on the vanadium ions (based on your calculations)?

Short Answer

Expert verified
a. 0.168 Faradays; b. The charge on the vanadium ions is 3 (\( \text{V}^{3+} \)).

Step by step solution

01

Convert Time to Seconds

Current is usually given in terms of seconds, so convert the time from hours to seconds: \[ 1.90 ext{ hours} = 1.90 imes 60 imes 60 = 6840 ext{ seconds} \]
02

Calculate Total Charge (Coulombs)

Calculate the total charge passed through the solution using the formula: \[ Q = It \]where \( I \) is current in amperes and \( t \) is time in seconds:\[ Q = 2.50 imes 6840 = 17100 ext{ coulombs} \]
03

Adjust for Electroplating Efficiency

The efficiency of the electroplating process is given as 95%, meaning only 95% of the charge effectively contributes to the plating. Adjust the charge for efficiency:\[ Q_{ ext{effective}} = 0.95 imes 17100 = 16245 ext{ coulombs} \]
04

Calculate Faradays Required

1 Faraday is approximately \( 96485 \) coulombs. Calculate the number of Faradays using:\[ ext{Faradays} = \frac{Q_{ ext{effective}}}{96485} \]\[ ext{Faradays} = \frac{16245}{96485} \approx 0.168 \text{ Faradays} \]
05

Calculate Moles of Vanadium Deposited

Convert the mass of vanadium deposited to moles using its molar mass (approximately 50.94 g/mol):\[ ext{Moles of V} = \frac{2.850}{50.94} = 0.056 \text{ moles} \]
06

Determine Charge on Vanadium Ions

The relationship between the charge passed, moles of substance, and faradays is given by:\[ ext{n Faradays} = ext{moles of substance} \times ext{valency} \]Using the calculated values: \[ 0.168 = 0.056 \times ext{valency} \]Solving for the valency:\[ ext{valency} = \frac{0.168}{0.056} = 3 \]
07

Verify the Result

The calculated charge on the vanadium ions is 3, which suggests the ion is likely \( ext{V}^{3+} \). This is consistent with known oxidation states of vanadium.

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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 concept in electrochemistry that relates the amount of electric charge used in an electrochemical reaction to the amount of substance produced or consumed.
It is based on the fact that the passage of a certain quantity of charge, known as one Faraday, will result in the oxidation or reduction of one mole of electrons. One Faraday is approximately equal to 96,485 coulombs.

In the context of electroplating, Faraday's Law helps us determine how much of a material will be deposited when a specific current is applied over a certain time period.
  • To apply Faraday's law, you begin with the total charge passed, which is the product of current (in amperes) and time (in seconds).
  • This charge is then adjusted for efficiency, as not all the charge may contribute effectively due to factors such as resistance and side reactions.
  • Finally, the effective charge is divided by the value of one Faraday to find the number of Faradays required for the process.
This law is not only useful in practical applications like electroplating but also aids in understanding the stoichiometry behind redox reactions in electrochemistry.
Vanadium Ions
Vanadium is a transition metal that can exist in multiple oxidation states, including +2, +3, +4, and +5. These oxidation states correspond to different charges on the vanadium ions. Understanding the charge or valency of vanadium ions is crucial for accurately performing electrochemical calculations.
In the given problem, the charge on the vanadium ions was calculated by using Faraday's Law, first determining the total amount of charge passed, then adjusting for electroplating efficiency. This effective charge is then used in conjunction with the known moles of vanadium deposited to find the valency of the vanadium ion.
  • Common vanadium ion: In this exercise, it was determined to be V3+, with a valency of +3.
  • Knowing the valency is important as it affects how the vanadium ions interact with electrodes during electrolysis.
  • The calculated valency must align with the known chemistry of vanadium ions, as seen in various industrial applications.
This information is pivotal for anyone working with vanadium in electrochemical processes, ensuring the correct form and behavior of the ions are accounted for.
Electroplating Efficiency
Electroplating involves coating a surface with a metal layer using an electric current. Efficiency in electroplating is vital as it indicates how much of the electrical energy is effectively used to deposit the desired metal.
Efficiency is expressed as a percentage, showing how much of the charge is successfully utilized in forming the metal layer. For example, if the efficiency is 95%, it means that 95% of the total charge contributes to the metal deposition.
  • An efficiency of less than 100% points to losses, possibly due to competing side reactions or resistance in the process.
  • In the case study, the efficiency affected the amount of vanadium deposited, emphasizing the need for optimizing conditions to reduce losses.
  • Calculation: To adjust for actual, effective charge used, multiply total charge by efficiency (as a decimal), which gives the charge that effectively contributes to plating.
High electroplating efficiency ensures the process is cost-effective and resource-efficient, reducing waste and operational costs while maximizing output.

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Most popular questions from this chapter

How many faradays are required for each of the following processes? How many coulombs are required? a. Reduction of \(1.0 \mathrm{~mol} \mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) b. Reduction of \(1.0 \mathrm{~mol} \mathrm{Fe}^{3+}\) to \(\mathrm{Fe}\) c. Oxidation of \(1.0 \mathrm{~g} \mathrm{Sn}^{2+}\) to \(\mathrm{Sn}^{4+}\) d. Reduction of \(1.0 \mathrm{~g} \mathrm{Au}^{3+}\) to \(\mathrm{Au}\)

In an analytical determination of arsenic, a solution containing arsenious acid, \(\mathrm{H}_{3} \mathrm{AsO}_{3}\), potassium iodide, and a small amount of starch is electrolyzed. The electrolysis produces free iodine from iodide ion, and the iodine immediately oxidizes the arsenious acid to hydrogen arsenate ion, \(\mathrm{HAsO}_{4}{ }^{2-}\). $$ \mathrm{I}_{2}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\underset{2 \mathrm{I}^{-}(a q)+\mathrm{HAsO}_{4}^{2-}(a q)+4 \mathrm{H}^{+}(a q)} $$ When the oxidation of arsenic is complete, the free iodine combines with the starch to give a deep blue color. If, during a particular run, it takes \(65.4 \mathrm{~s}\) for a current of \(10.5 \mathrm{~mA}\) to give an endpoint (indicated by the blue color), how many grams of arsenic are present in the solution?

Give the notation for a voltaic cell constructed from a hydrogen electrode (cathode) in \(1.0 M \mathrm{HCl}\) and a nickel electrode (anode) in \(1.0 \mathrm{M} \mathrm{NiSO}_{4}\) solution. The electrodes are connected by a salt bridge.

When molten lithium chloride, \(\mathrm{LiCl}\), is electrolyzed, lithium metal is liberated at the cathode. How many grams of lithium are liberated when \(5.00 \times 10^{3} \mathrm{C}\) of charge passes through the cell?

Balance the following skeleton equations. The reactions occur in acidic or basic aqueous solution, as indicated. a. \(\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{~S} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{S}_{8} \quad\) (acidic) b. \(\mathrm{Zn}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{N}_{2} \mathrm{O}\) (acidic) c. \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}\) (basic) d. \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-}\) (basic)
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