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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. |
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Answer» Solution :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` |
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