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?

Step by step solution

01

Calculate the volume of Ag used in one electrode

The surface area of the button is given as \(2.0 \, \text{cm}^2\) and the thickness of the silver layer is \(7.5 \times 10^{-6} \, \text{m}\). Convert the thickness to centimeters: \(7.5 \times 10^{-6} \, \text{m} = 7.5 \times 10^{-4} \, \text{cm}\). Now, use the formula for the volume of a rectangular prism to calculate the volume: \[ \text{Volume} = \text{Area} \times \text{Thickness} = 2.0 \, \text{cm}^2 \times 7.5 \times 10^{-4} \, \text{cm} = 1.5 \times 10^{-3} \, \text{cm}^3 \] The volume of Ag used in one electrode is \(1.5 \times 10^{-3} \, \text{cm}^3\).

02

Calculate the mass of Ag used per electrode

We are given the density of silver as \(10.5 \, \text{g/cm}^3\). Use the formula for mass: \[ \text{Mass} = \text{Density} \times \text{Volume} = 10.5 \, \text{g/cm}^3 \times 1.5 \times 10^{-3} \, \text{cm}^3 = 1.575 \times 10^{-2} \, \text{g} \] The mass of silver used per electrode is \(0.01575 \, \text{g}\).

03

Calculate the plating time

Given a current of \(12.0 \, \text{mA}\), first convert to amps: \(12.0 \, \text{mA} = 0.012 \, \text{A}\). The molar mass of Ag is \(107.87 \, \text{g/mol}\) and Faraday’s constant is \(96485 \, \text{C/mol}\). The number of moles of Ag required is: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.01575 \, \text{g}}{107.87 \, \text{g/mol}} \approx 1.460 \times 10^{-4} \, \text{mol} \] Calculate the total charge required: \[ \text{Charge} = \text{Moles} \times \text{Faraday's constant} = 1.460 \times 10^{-4} \, \text{mol} \times 96485 \, \text{C/mol} \approx 14.08 \, \text{C} \] Now, calculate the time in seconds using the formula \(Q = I \times t\): \[ t = \frac{Q}{I} = \frac{14.08 \, \text{C}}{0.012 \, \text{A}} \approx 1173.33 \, \text{s} \] Finally, convert seconds to minutes: \[ t = \frac{1173.33 \, \text{s}}{60} \approx 19.56 \, \text{minutes} \] The plating takes approximately \(19.56\) minutes.

04

Calculate the cost of silver in one electrode

Given the cost of silver as \(\$28.93\) per troy ounce (\(31.10 \, \text{g}\)), convert the cost to price per gram: \[ \text{Cost per gram} = \frac{28.93 \, \$}{31.10 \, \text{g}} \approx 0.9306 \, \$/\text{g} \] Now, calculate the cost of the silver used in one electrode: \[ \text{Cost} = 0.9306 \, \$ / \text{g} \times 0.01575 \, \text{g} \approx 0.01465 \, \$ = 1.465 \, \text{cents}\] The cost of the silver in one disposable electrode is approximately \(1.465\) cents.

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Key Concepts

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

Volume Calculation

First, let's understand how to calculate the volume of silver used in the electrode. Volume is a measurement of space in three dimensions. For the button, the surface area is given as 2.0 cm² and the thickness of the silver layer is 7.5 x 10⁻⁶ m. To calculate volume, convert thickness to centimeters: 7.5 x 10⁻⁶ m becomes 7.5 x 10⁻⁴ cm.
By using the volume formula for a rectangular prism, we have:
\text{Volume} = \text{Area} \times \text{Thickness}
So,
\[ \text{Volume} = 2.0 \; \text{cm}^2 \times 7.5 \times 10^{-4} \; \text{cm} = 1.5 \times 10^{-3} \; \text{cm}^3 \]
The volume of silver used in one electrode equals 1.5 × 10⁻³ cm³.

Mass Calculation

Next, let's focus on how to calculate the mass of silver used per electrode. We use the known volume and the density of silver. Density is given as 10.5 g/cm³. The formula for mass is:
\text{Mass} = \text{Density} \times \text{Volume}
Substituting the known values:
\[ \text{Mass} = 10.5 \; \text{g/cm}^3 \times 1.5 \times 10^{-3} \; \text{cm}^3 = 1.575 \times 10^{-2} \; \text{g} \]
The mass of silver used per electrode is 0.01575 g. This can usually be understood as converting how much space the silver takes up into how much it weighs.

Plating Time

To calculate the time required for plating, we need to use the principles of electrochemistry. Given a current of 12.0 mA (or 0.012 A), and knowing that the molar mass of Ag is 107.87 g/mol, we first calculate the number of moles of Ag required:
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.01575 \; \text{g}}{107.87 \; \text{g/mol}} \approx 1.460 \times 10^{-4} \; \text{mol}
We then use Faraday’s constant (96485 C/mol) to calculate the total charge required:
\text{Charge} = \text{Moles} \times \text{Faraday's constant} = 1.460 \times 10^{-4} \; \text{mol} \times 96485 \; \text{C/mol} \approx 14.08 \; \text{C}
Finally, we use the formula \text{Q} = \text{I} \times \text{t}\to find the time in seconds:
\[ \text{t} = \frac{14.08 \; \text{C}}{0.012 \; \text{A}} \approx 1173.33 \; \text{s} \]
Converting this to minutes:
\[ \text{t} = \frac{1173.33 \; \text{s}}{60} \approx 19.56 \; \text{minutes} \]
The plating process takes approximately 19.56 minutes.

Cost Calculation

Finally, we calculate the cost of the silver used in one electrode. Given the cost of silver as $28.93 per troy ounce (31.10 g), we first find the cost per gram:
\[ \text{Cost per gram} = \frac{28.93 \; \$$}{31.10 \; \text{g}} \approx 0.9306 \; \$$/\text{g} \]
We then calculate the total cost of the silver in one electrode:
\[ \text{Cost} = 0.9306 \; \$$/\text{g} \times 0.01575 \; \text{g} \approx 0.01465 \; \$$ \]
Converting this to cents:
\[ \text{Cost} = 1.465 \; \text{cents} \]
The cost of the silver in one disposable electrode is approximately 1.465 cents.