An inexpensive and accurate method of measuring the quantity of electricity flowing through a circuit is to pass the current through a solution of a metal ion and weigh the metal deposited. A silver electrode immersed in an Ag \(^{+}\) solution weighs \(1.7854 \mathrm{~g}\) before the current has passed and weighs \(1.8016 \mathrm{~g}\) after the current has passed. How many coulombs have passed?

Short Answer

Expert verified
14.50 C

Step by step solution

01

- Calculate the mass of silver deposited

First, determine the mass of the silver deposited by subtracting the initial mass of the electrode from the final mass. Mass deposited = final mass - initial mass = 1.8016 g - 1.7854 g = 0.0162 g
02

- Convert mass to moles

Next, convert the mass of silver deposited to moles using the molar mass of silver (Ag), which is 107.87 g/mol. Number of moles = mass / molar mass = 0.0162 g / 107.87 g/mol ≈ 1.502 x 10^-4 mol
03

- Use Faraday's Law of Electrolysis

According to Faraday's law, the amount of substance deposited is directly proportional to the amount of electricity that passes through the solution. For silver, 1 mole of Ag is deposited by 1 mole of electrons (or 1 Faraday), which is equal to 96485 coulombs. Therefore, the number of coulombs passed can be calculated by multiplying the number of moles of Ag by 96485 C/mol: Coulombs = number of moles x Faraday constant = 1.502 x 10^-4 mol x 96485 C/mol ≈ 14.50 C

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Key Concepts

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

Coulombs Measurement
Coulombs measure the quantity of electrical charge that flows in a circuit. In electrochemistry, understanding how many coulombs have passed through a solution helps us know how much electric current has been used.
To calculate the coulombs from a known mass of deposited material, the mass is first measured before and after the current flows.
This mass difference helps us find the number of moles of the substance deposited.
Once we know the number of moles, we multiply it by Faraday's constant (96485 C/mol) to find the total coulombs.
In this exercise, you found a silver electrode’s mass change, converting it to moles, then calculating the total charge in coulombs.
Mass to Moles Conversion
First, we must convert the mass of deposited material into moles. Moles are a key unit in chemistry that measures the amount of substance. This conversion is essential since it helps link mass to the number of atoms or molecules.
Here's a step-by-step to convert mass to moles:
  • Measure the change in mass before and after the current passes.
  • Use the molar mass (from the periodic table) of the substance. In this example, we used silver.
  • Divide the mass of the substance by its molar mass.
For instance, the molar mass of silver is 107.87 g/mol. By dividing the mass change of silver by its molar mass, you get the number of moles deposited.
E.g., If 0.0162 g of silver was deposited, divide 0.0162 g by 107.87 g/mol (molar mass) to get around 1.502 x 10^-4 moles of silver.
Electrochemistry Calculations
Electrochemistry involves the study of the movement of electrons in chemical processes. Faraday's Law of Electrolysis plays a pivotal role.
This law states that the amount of any substance deposited or dissolved at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.
Using the law:
  • First, convert the deposited mass into moles.
  • Next, apply Faraday's constant (96485 C/mol). This constant links moles of electrons to electric charge.
  • Multiply the moles of the substance by 96485 C/mol to get the total charge in coulombs.
For example, you've converted 0.0162 g of silver to moles of silver (about 1.502 x 10^-4 moles). Multiply this by 96485 C/mol to find out how many coulombs have passed through the solution, resulting in approximately 14.50 C.
Understanding these calculations is fundamental for interpreting results and solving complex electrochemical problems.

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Most popular questions from this chapter

Electrodes used in electrocardiography are disposable, and many of them incorporate silver. The metal is deposited in a thin layer on a small plastic "button," and then some is converted to AgCl: $$\operatorname{Ag}(s)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \operatorname{AgCl}(s)+\mathrm{e}^{-}$$ (a) If the surface area of the button is \(2.0 \mathrm{~cm}^{2}\) and the thickness of the silver layer is \(7.5 \times 10^{-6} \mathrm{~m},\) calculate the volume (in \(\mathrm{cm}^{3}\) ) of \(\mathrm{Ag}\) used in one electrode. (b) The density of silver metal is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many grams of silver are used per electrode? (c) If \(\mathrm{Ag}\) is plated on the button from an \(\mathrm{Ag}^{+}\) solution with a current of \(12.0 \mathrm{~mA}\), how many minutes does the plating take? (d) If bulk silver costs \(\$ 28.93\) per troy ounce \((31.10 \mathrm{~g}),\) what is the cost (in cents) of the silver in one disposable electrode?

A chemist designs an ion-specific probe for measuring \(\left[\mathrm{Ag}^{+}\right]\) in an \(\mathrm{NaCl}\) solution saturated with \(\mathrm{AgCl}\). One half-cell has an Ag wire electrode immersed in the unknown AgCl-saturated \(\mathrm{NaCl}\) solution. It is connected through a salt bridge to the other half-cell, which has a calomel reference electrode [a platinum wire immersed in a paste of mercury and calomel (Hg \(_{2} \mathrm{Cl}_{2}\) ) ] in a saturated KCl solution. The measured \(E_{\text {cell }}\) is \(0.060 \mathrm{~V}\). (a) Given the following standard half-reactions, calculate \(\left[\mathrm{Ag}^{+}\right]\). Calomel: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Hg}(I)+2 \mathrm{Cl}^{-}(a q) \quad E^{\circ}=0.24 \mathrm{~V}\) Silver: \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)\) $$E^{\circ}=0.80 \mathrm{~V}$$ (Hint: Assume that [Cl \(^{-}\) ] is so high that it is essentially constant.) (b) A mining engineer wants an ore sample analyzed with the \(\mathrm{Ag}^{+}-\) selective probe. After pretreating the ore sample, the chemist measures the cell voltage as \(0.53 \mathrm{~V}\). What is \(\left[\mathrm{Ag}^{+}\right] ?\)

When zinc is refined by electrolysis, the desired halfreaction at the sathnde is $$\mathrm{Zn}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(s)$$ A competing reaction, which lowers the yield, is the formation of hydrogen gas: $$2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g)$$ If \(91.50 \%\) of the current flowing results in zinc being deposited, while \(8.50 \%\) produces hydrogen gas, how many liters of \(\mathrm{H}_{2},\) measured at STP, form per kilogram of zinc?

A voltaic cell consists of two \(\mathrm{H}_{2} / \mathrm{H}^{+}\) half-cells. Half-cell \(\mathrm{A}\) has \(\mathrm{H}_{2}\) at 0.95 atm bubbling into \(0.10 \mathrm{M} \mathrm{HCl}\). Half-cell \(\mathrm{B}\) has \(\mathrm{H}_{2}\) at 0.60 atm bubbling into \(2.0 \mathrm{M} \mathrm{HCl}\). Which half-cell houses the anode? What is the voltage of the cell?

Bubbles of \(\mathrm{H}_{2}\) form when metal \(\mathrm{D}\) is placed in hot \(\mathrm{H}_{2} \mathrm{O}\). No reaction occurs when \(\mathrm{D}\) is placed in a solution of a salt of metal E, but D is discolored and coated immediately when placed in a solution of a salt of metal \(\mathrm{F}\). What happens if \(\mathrm{E}\) is placed in a solution of a salt of metal F? Rank metals \(D, \mathrm{E},\) and \(\mathrm{F}\) in order of increasing reducing strength.

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