Q21.130 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Magnesium bars are connected electrically to underground iron pipes to serve as sacrificial anodes. 

(a) Do electrons flow from the bar to the pipe or the reverse? 

(b) A$\text{12kg Mg}$ bar is attached to an iron pipe, and it takes  $\text{8.5 yr}$ for the Mg to be consumed. What is the average current flowing between the Mg and the Fe during this period?

Step-by-Step Solution

Answer

(a) The electrons flow from bar to pipe.

(b) The average current flowing between Mg and the Fe is 0.36A .

Step 1: Concept Introduction

An electrolytic cell is an electrochemical device that employs electrical energy to help a non-spontaneous redox reaction take place. Electrolytic cells are electrochemical cells that can be used to electrolyze a variety of substances.

Step 2: (a) Flow of Electrons

Mg bars serve as sacrificial anodes; therefore, oxidation takes place at the Mg bars. Electrons thus move from the Mg bars to the iron pipes.

 

Therefore, electrons go towards the pipe from the bar.

Step 3: (b) Calculation for Current

To compute the current, get the charge and divide it by the time.

Convert the given mass of Mg to moles of Mg .

$\text{12kg×}\frac{\text{1000g}}{\text{1kg}}\text{×}\frac{\text{1mol Mg}}{\text{24.31g}}\text{ = 493.62mol Mg}$

Convert the moles of Mg to moles of electrons. There are 2 electrons involved in the reduction of ${\text{Mg}}^{\text{2 + }}\to \text{Mg}$ .

$\text{493.62mol Mg×}\frac{{\text{2mol e}}^{\text{ - }}}{\text{1mol Mg}}{\text{ = 987.24mol e}}^{\text{ - }}$

Multiply by Faraday's constant to get the charge.

${\text{987.24 mol }}^{\text{ - }}\text{×}\frac{\text{96500 C}}{{\text{mol e}}^{\text{ - }}}\text{ = 9.53}·{\text{10}}^{\text{7}}\text{C}$

Convert the given amount of time from years to seconds.

$\text{8.5 years×}\frac{\text{365.25 days }}{\text{ year }}\text{×}\frac{\text{24 hours }}{\text{1 day }}\text{×}\frac{\text{60min}}{\text{ hour }}\text{×}\frac{\text{60 seconds }}{\text{min}}\text{ = 2.682}·{\text{10}}^{\text{8}}\text{ seconds}$

Divide the charge by the time to get the current.

$\frac{\text{9.53}·{\text{10}}^{\text{7}}\text{C}}{\text{2.682}·{\text{10}}^{\text{8}}\text{s}}\text{ = 0.36A}$

 

Therefore, the value for current is obtained as $0.36A$.