Question

MgCl₂. The reaction at the cathode is:

Mg²+(l) + 2e⁻ → Mg(l).


how many grams of magnesium metal are produced if an average current of 66.7 A flows for 4.45hr? assume all of the car is consumed by the half reaction shown.

Answer 1

268.6 grams of magnesium metal are produced if an average current of 66.7A flows for 4.45hr in the given reaction.

Step-by-step explanation:

The reaction at the cathode is represented by the half-reaction: Mg²+(l) + 2e⁻ → Mg(l). To calculate the grams of magnesium metal produced, we need to determine the amount of charge that flowed during the electrolysis.

Using the formula Q = I * t, where I is the current and t is the time in seconds, we can convert the given current of 66.7 A for 4.45 hours into seconds and calculate the total charge.

Then we can use Faraday's constant (F = 96,485 C/mol) and the molar mass of magnesium (24.31 g/mol) to convert the charge into moles of Mg²+.

Finally, using the mole ratio from the balanced equation (1:1), we can determine the grams of magnesium produced.

First, convert the time from hours to seconds:

4.45 hours * 3600 seconds/hour

= 16020 seconds

Next, calculate the total charge (Q) using the formula:

Q = I * t = 66.7 A * 16020 s

= 1,067,340 C

Convert the total charge to moles of Mg²+ using Faraday's constant:

moles of Mg²+ = Q / F = 1,067,340 C / 96,485 C/mol

= 11.06 mol

Finally, calculate the grams of magnesium produced using the molar mass of magnesium:

grams of Mg = moles of Mg²+ * molar mass of Mg

= 11.06 mol * 24.31 g/mol

= 268.6 g.