Q21.119 CP - Chemistry: Molecular Nature Of Matter And Change

Question

A silver button battery used in a watch contains $\text{0.75 g}$ of zinc and can run until $\text{80\% }$ of the zinc is consumed. 

(a) How many days can the battery run at a current $\text{0.85}$ of  microamps (${\text{10}}^{\text{ - 6}}\text{ amps}$)? 

(b) When the battery dies, $\text{95\% }$ of the ${\text{Ag}}_{\text{2}}\text{O}$ has been consumed. How many grams of  was used to make the battery? 

(c) If  Ag costs $\text{\$ 13.00}$ per troy ounce ($\text{31.10 g}$), what is the cost of the  consumed each day the watch runs?

Step-by-Step Solution

Answer

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.

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 Time

• The reaction is $\text{Zn(s)}\to {\text{Zn}}^{\text{2 + }}{\text{(aq) + 2e}}^{\text{ - }}$
• The mass of $\text{Zn}$ is $\text{0.75 g}$ .
• The mass of $\text{Zn}$consumed, until the battery dies, is ${\text{m}}_{\text{Zn}}\text{ = 0.75 g}·\text{80\% = 0.6 g}$
• The current is $\text{0.85μA=0.85}·{\text{10}}^{-6}\text{A(A = C/s)}$
• The ideal gas constant is $\text{R = 0.0821 L atm/K mol}$
• Faraday constant: charge of 1 of electrons ($\text{F = 96485 C/mole}$)

The number of moles of $\text{Zn}$ is (molar mass $\text{Zn}$of $\text{65.38g/mol}$  is ) –

$$\begin{gathered} {\text{n}}_{\text{Zn}}\text{ = 0.6g}\frac{\text{1mol}}{\text{65.38g/mol}} \\ \text{ = 9.18}·{\text{10}}^{\text{ - 3}}\text{mol} \end{gathered}$$

Since  2 moles of electrons are involved in consumption of  1 mole of Zn , the number of moles of electrons is –

$$\begin{gathered} {\text{n}}_{e^{-}}\text{=9.18}·{\text{10}}^{-3}\text{mol Zn}\frac{{\displaystyle {\text{2 mol e}}^{-}}}{{\displaystyle \text{1 mol Zn}}} \\ \text{ = 1.836}·{\text{10}}^{-2}{\text{mol e}}^{-} \end{gathered}$$

Calculate the charge using Faraday constant –

$$\begin{gathered} {\text{Charge = Moles e}}^{\text{ - }}·\text{F} \\ \text{ = 1.836}·{\text{10}}^{\text{ - 2}}{\text{mol e}}^{\text{ - }}·{\text{96485 C/mol e}}^{\text{ - }} \\ \text{ = 1.77}·{\text{10}}^{\text{3}}\text{ C} \end{gathered}$$

The time it takes to deposit the gold on one side of one earring:

$$\begin{gathered} \text{Current = }\frac{\text{ Charge }}{\text{ Time }} \\ \text{Time = }\frac{\text{ Charge }}{\text{ Current }} \\ \text{ = }\frac{\text{1.77}·{\text{10}}^{\text{3}}\text{ C}}{\text{0.85}·{\text{10}}^{\text{ - 6}}\frac{\text{C}}{\text{s}}} \\ \text{ = 2.08}·{\text{10}}^{\text{9}}\text{s} \\ \text{ = 2.08}·{\text{10}}^{\text{9}}\text{s}·\frac{\text{1h}}{\text{3600s}}·\frac{\text{1d}}{\text{24h}} \\ \text{ = 2.4}·{\text{10}}^{\text{4}}\text{ days} \end{gathered}$$

Therefore, the vale for number of days is obtained as $\text{2.4}·{\text{10}}^{\text{4}}\text{ days}$ .

Step 3: (b) Calculation for Mass

• The reaction is –
• When battery dies, $\text{95\% }$ of  ${\text{Ag}}_{\text{2}}\text{O}$ is consumed.
• The number of moles of electrons is $\text{1.836}·{\text{10}}^{\text{ - 2}}\text{mol}$ .

Since  of electrons is involved in consumption 1 mole of Ag , the number of moles of  Ag is $\text{1.836}·{\text{10}}^{\text{ - 2}}$.

Now, we will calculate the mass of Ag consumed, when battery died (molar mass of Ag is $\text{107.8682 g/mol}$ ) –

$$\begin{gathered} {\text{m}}_{\text{Ag}}\text{ consumed = 1.836}·{\text{10}}^{\text{ - 2}}\text{mol}·\text{107.8682g/mol} \\ \text{ = 1.98g} \end{gathered}$$

Since this represents  $\text{95\% }$of the original mass of $\text{Ag}$ , the mass of $\text{Ag}$ in a battery is –

$$\begin{gathered} {\text{m}}_{\text{Ag}}\text{ = }\frac{\text{1.98g}}{\text{95\% }}·\text{100\% } \\ \text{ = 2.08g} \end{gathered}$$

Therefore, the value for mass is obtained as $\text{2.08g}$

 

Step 3: (c) Calculation for Cost

$\text{31.10 g}$ of $\text{Ag}$ costs  $\text{13.00}$dollars

The cost of $\text{Ag}$ in a battery is –

$$\begin{gathered} \text{Cost of Ag in battery = 2.08g}·\frac{\text{13.00 dollars }}{\text{31.10g}} \\ \text{ = 0.869 dollars/battery} \end{gathered}$$

Now, calculate the cost of $\text{Ag}$ in a battery per day –

$$\begin{gathered} \text{Cost per day = 0.869 dollars/battery}·\frac{\text{1 battery }}{\text{2.4}·{\text{10}}^{\text{4}}\text{d}} \\ \text{ = 3.62}·{\text{10}}^{\text{ - 5}}\text{dollars/d} \end{gathered}$$

 

Therefore, the value for cost is obtained as $\text{3.62}·{\text{10}}^{\text{ - 5}}\text{dollars/d}$ .