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A \(100.0 \mathrm{~mL}\) dilute solution of \(\mathrm{Ag}^{+}\) is electrolyzed for \(15.0\) minutes with a current of \(1.25 \mathrm{~mA}\) and the silver is removed completely. What was the initial \(\left[\mathrm{Ag}^{+}\right]\) ? (a) \(2.32 \times 10^{-1}\) (b) \(2.32 \times 10^{-4}\) (c) \(2.32 \times 10^{-3}\) (d) \(1.16 \times 10^{-4}\)

Short Answer

Expert verified
\(2.32 \times 10^{-3}\) M

Step by step solution

01

Calculate the total charge passed through the solution

To calculate the total charge (Q) that passed through the solution, use the formula Q = It, where I is the current and t is the time the current flows. Convert the current from milliamperes (mA) to amperes (A) and the time from minutes to seconds before multiplying them together.
02

Determine the amount of silver deposited

Using Faraday's laws of electrolysis, the amount of silver (n) deposited is related to the charge by the equation n = Q / F, where F is Faraday's constant (approximately 96485 C/mol). Since the charge on one mole of \(\mathrm{Ag}^+\) ions is equivalent to one Faraday, the charge Q will directly give us the moles of \(\mathrm{Ag}^+\) deposited.
03

Calculate initial concentration of \(\mathrm{Ag}^+\)

The initial concentration \([\mathrm{Ag}^+]\) can be found by dividing the number of moles of \(\mathrm{Ag}^+\) by the volume of the solution. Convert the volume from milliliters (mL) to liters (L) before calculating the concentration.
04

Express the answer in scientific notation

To match the given answer format, express the calculated concentration in terms of scientific notation.

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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 fundamental to understanding the quantitative aspects of electrolysis. Electrolysis is a process where electrical energy is used to drive a non-spontaneous chemical reaction. It's a technique commonly used in electroplating, refining metals, and for applications such as producing hydrogen gas.

Faraday's first law states that the amount of a substance altered at an electrode during electrolysis is directly proportional to the amount of electricity that passes through the circuit. This principle helps us quantify the changes occurring during the electrolytic process. The second law points out that for the same quantity of electricity passed through different electrolytes, the amount of substances liberated is proportional to their equivalent weights.

These laws help us calculate things like how much metal is deposited during electroplating. For instance, in our example problem, Faraday's first law allows us to determine the moles of silver deposited by using the charge that passed through the solution and dividing it by Faraday's constant. Understanding these laws provides the foundation to approach and solve quantitative electrolysis problems effectively.
Calculating Concentration
Determining the concentration of a solution is a pivotal skill in chemistry. Concentration refers to the amount of a substance in a defined space. It can be expressed in various units such as molarity, which is the number of moles of solute per liter of solution.

To calculate the concentration, you need to know the amount of solute (in moles) and the volume of the solution. In the context of our electrolysis problem, after using Faraday's law to find the moles of silver deposited, we then calculate the initial concentration of silver ions by dividing these moles by the total volume of the solution in liters.

Why are these calculations crucial? They are used in preparing solutions for reactions, pharmaceutical formulations, and any chemical process where the exactness of reactant concentration is necessary for consistency and desired outcomes. Through practice, students can master this basic and essential technique, enhancing their lab competencies and their grasp of the underlying principles.
Quantitative Aspects of Electrolysis
The quantitative aspects of electrolysis revolve around calculating the amounts of substances involved in the electrochemical reaction. It combines Faraday's laws with basic stoichiometry. For example, when current is passed through an electrolytic cell, knowing the amount of charge and the duration allows us to calculate the amount of substance that will be deposited at the electrode.

In our specific exercise, this involves calculating the total charge passed (using the current and the time), then relating that charge to the number of moles of silver deposited. Once the moles are known, the initial concentration before electrolysis can be calculated. This process shows the interconnection between charge, moles, and concentration, all of which are key to understanding and solving problems related to electrolysis.

Gaining a solid grasp of these quantitative relationships not only aids in solving textbook problems but also prepares students for practical applications in industries where precise chemical deposition is required, thereby highlighting the vital role of electrolysis in various scientific and engineering fields.

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Most popular questions from this chapter

In electrochemical corrosion of metals, the metal undergoing corrosion : (a) acts as anode (b) acts as cathode (c) undergoes reduction (d) None

Based on the following information arrange four metals \(A, B, C\) and \(D\) in order of decreasing ability to act as reducing agents: (I) Only \(A, B\) and \(C\) react with \(1 M \mathrm{HCl}\) to give \(\mathrm{H}_{2}(g)\) (II) When \(C\) is added to solutions of the other metal ions, metallic \(B\) and \(D\) are formed (III) Metal \(C\) does not reduce \(A^{n+}\). (a) \(C>A>B>D\) (b) \(C>A>D>B\) (c) \(A>C>D>B\) (d) \(A>C>B>D\)

Zn can displace : (a) \(\mathrm{Mg}\) from its aqueous solution (b) Cu from its aqueous solution (c) Na from its aqueous solution (d) Al from its aqueous solution

When a solution of \(\mathrm{AgNO}_{3}(1 \mathrm{M})\) is electrolyzed using platinum anode and copper cathode. What are the products obtained at two electrodes? Given : \(E_{\mathrm{Cu}^{2+}}^{\circ} \mathrm{Cu}=+0.34\) volt; \(\quad E_{\mathrm{O}_{2}, \mathrm{H}^{+} \mid \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.23\) volt \(; \quad E_{\mathrm{H}^{+} \mid \mathrm{H}_{2}}^{\circ}=+0.0\) volt; \(E_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^{\circ}=+0.8 \mathrm{volt}\) (a) \(\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}\) at anode; \(\mathrm{Ag}^{+} \longrightarrow \mathrm{Ag}\) at cathode (b) \(\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{O}_{2}\) at anode; \(\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}\) at cathode (c) \(\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{O}_{2}\) at anode; \(\mathrm{Ag}^{+} \longrightarrow \mathrm{Ag}\) at cathode (d) \(\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2}\) at anode; \(\mathrm{Ag}^{+} \longrightarrow \mathrm{Ag}\) at cathode

Consider the following equations for a cell reaction $$ \begin{gathered} A+B \rightleftharpoons C+D ; \quad E^{\circ}=x \text { volt, } K_{e q}=K_{1} \\\ 2 A+2 B \rightleftharpoons 2 C+2 D ; E^{\circ}=y \text { volt, } K_{e q}=K_{2} \end{gathered} $$ then : (a) \(x=y, K_{1}=K_{2}\) (b) \(x=2 y, K_{1}=2 K_{2}\) (c) \(x=y, K_{1}^{2}=K_{2}\) (d) \(x^{2}=y, K_{1}^{2}=K_{2}\)

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