Q21.123 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Electrodes used in electrocardiography are disposable, and many incorporate silvers. The metal is deposited in a thin layer on a small plastic “button,” and then some is converted to $\text{AgCl}$: 

${\text{Ag(s) + Cl}}^{\text{ - }}\text{(aq)}\rightleftharpoons {\text{AgCl(s) + e}}^{\text{ - }}$

(a) If the surface area of the button is ${\text{2.0 cm}}^{\text{2}}$ and the thickness of the silver layer is $7.5\times {10}^{-6}m$, calculate the volume (in ${\text{cm}}^{\text{3}}$) of  used in one electrode. 

(b) The density of silver metal is ${\text{10.5 g/cm}}^{\text{3}}$. How many grams of silver are used per electrode? 

(c) If  is plated on the button from an ${\text{Ag}}^{\text{ + }}$ solution with a current of $\text{12.0 mA}$ , how many minutes does the plating take? 

(d) If bulk silver costs  $\text{\$ 13.00}$per troy ounce ($\text{31.10 g}$), what is the cost (in cents) of the silver in one disposable electrode?

Step-by-Step Solution

Answer

(a) The volume of $\text{Ag}$ used in one electrode is $\text{1.5}·{\text{10}}^{\text{ - 3}}{\text{ cm}}^{\text{3}}$ .

(b) The amount of silver used per electrode is $\text{0.016 g}$ .

(c) The plating takes the time of $\text{19.57min}$.

(d) The cost of the silver in one disposable electrode is $\text{0.66 cents}$ .

Step 1: Concept Introduction

The process of dissolving ionic compounds into their constituent components by delivering a direct electric current through the complex in a fluid form is known as electrolysis. At the cathode, cations are reduced, whereas anions are oxidised.

Step 2: (a) Calculation for Volume

To get the volume of  used, multiply the surface area by the thickness.

Convert the thickness to .

$\text{7.5}·{\text{10}}^{\text{ - 6}}\text{m×}\frac{\text{100 cm}}{\text{1 m}}{\text{ = 7.5.10}}^{-4}\text{cm}$

Multiply the thickness by the area.

$$\begin{gathered} {\text{V = 2.0 cm}}^{\text{2}}{\text{×75.10}}^{-4}\text{cm} \\ =\text{1.5}·{\text{10}}^{-3} c{m}^3 \end{gathered}$$

Therefore, the value for volume is obtained as .

Step 3: (b) Calculation for Mass

To get the mass of $\text{Ag}$, multiply the volume by the density.

$$\begin{gathered} \text{m = 1.5}·{\text{10}}^{-3}\text{×}\frac{\text{10.5 g}}{{\text{cm}}^{\text{3}}} \\ \text{ = 0.016 g Ag} \end{gathered}$$

 

Therefore, the value for mass is obtained as $\text{0.016 g}$ .

Step 4: (c) Calculation for Time

Get the time given the mass and the current.

Convert the given mass to mol by multiplying by the molar mass of $\text{Ag}$ .

$\text{0.016 g Ag×}\frac{\text{1mol Ag}}{\text{107.9 g Ag}}\text{ = 1.46}·{\text{10}}^{-4}\text{mol Ag}$

Convert it to moles of electrons, and there's one electron involved in the conversion of .

${\text{Ag}}^{\text{ + }}\to \text{Ag}$

$\text{1.46}·{\text{10}}^{-4}\text{mol Ag×}\frac{\text{1mole}}{\text{1mol Ag}}\text{ = 1.46}·{\text{10}}^{-1}{\text{mol e}}^{-}$

Multiply the Faraday's constant, and divide by the current to get the time.

$$\begin{gathered} \text{1.46}·{\text{10}}^{-4}{\text{mol e}}^{-}\text{×}\frac{{\displaystyle \text{96500 C}}}{{\displaystyle {\text{1mol e}}^{-}}}\text{×}\frac{\text{1s}}{\text{12.0 m C}}\text{×}\frac{\text{1 mC}}{\text{0.001 mC}}\text{×}\frac{\text{1min}}{\text{60s}} \\ \text{t = 19.57min} \end{gathered}$$

 Therefore, the value for time is obtained as $\text{19.57min}$.

Step 5: (d) Calculation for Cost

Multiply the mass by the cost to get the price. Multiply by 100 to convert it to cents.

 $$\begin{gathered} \text{0.016 g×}\frac{\text{13.00 USD}}{\text{31.10 g}}\text{×}\frac{\text{100 cents}}{\text{1 USD}} \\ \text{cost = 0.66 cents} \end{gathered}$$

Therefore, the value for cost is obtained as $\text{0.66 cents}$ .