InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Calculalting the amount the product from the amount of charge in an electrolysis: A constant current of `0.452 A` is passed through an electrolytic cell containing molten `CaCl_(2)` for a time of `1.50` hours. Write the electrode reactions and calculate the quantity of products (in grams) formed at the electrodes. Also find the volume (at STP) of any gaseous product formed. Strategy: to convert the current and time to grams or litres of product, carry out the sequence of conversions in Figure 3.9. |
|
Answer» Since the only ions present in molten `CaCl_(2)` are `Ca^(2+)` and `Cl^(-)`, the reactions are `{:("Anode": 2Cl^(-)(1)rarr Cl_(2)(g)+2e^(-)),("Cathode": Ca^(2+)(1)+2e^(-)rarrCa(1)),(bar("Overall": Ca^(2+)(1)+2Cl^(-)(1)rarr Ca(1)+Cl_(2)(g))):}` The quantities of `Ca` metal and `Cl_(2)` gas formed depend on the number of electrons that pass through the electrolytic cell, which in turn depends on the current and time or charge. Step 1: Because electrons can be thought of as a reactant in the electrolysis process, the first step is to calculate the charge and the number of moles of electrons passed throgh the cell: `Q = It` Charge `= (0.452 A)(1.50h)((3600s)/(1h))((1 C)/(1 A.s))` `(0.452(C)/(s)) (1.50h) ((60 min)/(h)) ((60 s)/(min))` `= 2440.8 C = 2.44 xx 10^(3)C` Moles of `e^(-) = 2.44 xx 10^(3)C (("1 mol" e^(-))/(96,500C))` `= 0.025 mol e^(-)` Step 2 : The cathode reaction yield `1` mol of `Ca` per `2` mol of electrons, so `0.025//2` or `0.0125` mol of `Ca` will be obtained: `Ca^(2+) + 2e^(-) rarr Ca` Moles of `Ca = (0.025 mol e^(-)) ((1 mol Ca)/(2 mol e^(-)))` `= 0.0125 mol Ca` Step 3: Converting the number ofmoles of `Ca` to grams of `Ca` gives Grams of `Ca = (0.0125 mol Ca) ((40g Ca)/(mol Ca))` `= 0.5 g Ca` As a shortcut, the entire porcess of conversion of coulombs to grams can be carried out in one step: `? g Ca = (2.44 xx 10^(3)C)(1 "mole"^(-))/(96,500 C)(1 mol Ca)/(2 "mole"^(-))((40 g Ca)/(1 mol Ca))` `= 0.5 g Ca` Step 4: The anode reaction gives `1 mol` of `Cl_(2)` per 2 mol of electrons, so `0.0125 mol` of `Cl_(2)` will be obtained: `2Ci^(-)(l) rarr Cl_(2)(g) + 2e^(-)` Moles of `Cl_(2) = (0.025 "mole"^(-)) ((1 mol Cl_(2))/(2 mol E^(-)))` `= 0.0125 mol Cl_(2)` Step 5: Converting the number of moles of `Cl_(2)` to grams of `Cl_(2)` gives Grams of `Cl_(2) = (0.125 mol Cl_(2)) ((71 g Cl_(2))/(1 mol Cl_(2)))` `= 8.88 g Cl_(2)` Step 6: Since `1` mole of an ideal gas occupies `22.4 L` at STP, the volume of `Cl_(2)` obtained is Litres of `Cl_(2) = (0.125 mol Cl_(2))((22.4 L Cl_(2))/(1 mol Cl_(2)))` `= 0.28 L` As a shortcut, the entire sequence of conversions can be carried out in just one step. For example the volume of `Cl_(2)` produced at the anode is `(0.452(C)/(s))(1.50 h)((3600s)/(h))((1mol e^(-))/(96,500C))((1molCl_(2))/(2 "mole"^(-)))` `((22.4LCl_(2))/(1molCl_(2)) = 0.28 L` |
|