Claculate the mass of chromium that can be deposited by electrolysis of an aqueous solution of chromium(III)sulfate,Cr₂(SO₄)₃,for 180 min using a constant current of 11.0 A.Assume 100% current efficiency.(F = 96485 C/mol) A) 0.356 g B) 21.3 g C) 192.1 g D) 0.187 g E) 32.0 g

Question

Quizplus · Accepted Answer

First, we need to calculate the total charge passed through the solution using the formula:
Q = I * t
Where I is the current (11.0 A) and t is the time in seconds (180 min = 10800 s)
Q = 11.0 A * 10800 s = 118800 C
Next, we need to convert the charge to moles of electrons using Faraday's constant:
n = Q / F
n = 118800 C / 96485 C/mol = 1.23 mol e-
Since the electrolysis of chromium(III) sulfate involves the reduction of the Cr3+ ion to chromium metal, we know that one mole of electrons will produce one mole of chromium:
Cr3+ + 3e- → Cr
So we can say that 1.23 mol of electrons will produce 1.23 mol of chromium.
Finally, we can calculate the mass of chromium using its molar mass:
Molar mass of chromium = 52.0 g/mol
Mass of chromium = 1.23 mol * 52.0 g/mol = 64.0 g
Therefore, the mass of chromium that can be deposited is approximately 64.0 g. However, we need to take into account the assumption of 100% current efficiency. In reality, there will be some losses due to side reactions and other factors. Therefore, the actual mass of chromium deposited will be less than 64.0 g. The answer closest to this value is B) 21.3 g.

Claculate the Mass of Chromium That Can Be Deposited by Electrolysis