Chromium is reduced from CrO4{2-} to Cr(OH)4- in an electrolytic cell. What mass of Cr(OH)4- is formed by passage of 4.50 A for a period of 8000 seconds?
Isn't an amp a coulomb of electrons?
number electrons=1coulomb/charge on each electron
Now, after you figure that out, multiply it by 4.5*8000 to get the number of electrons.
Thence, divide it by avagradros number to convert it to moles of electrons.
It appears the chronium went from +6 to +5, or each ion taking on one electron.
That number of moles of electrons is the same number of moles of CrO4 -2 ions, so what is the mass of that?
To calculate the mass of Cr(OH)4- formed, we need to use the concepts of Faraday's laws and stoichiometry.
First, let's determine the number of moles of electrons transferred during the electrolysis process. We know that one mole of electrons corresponds to the charge of 1 Faraday, which is 96,485 C.
Given that the current is 4.50 A and the time is 8000 seconds, we can calculate the total charge passed through the cell using the equation:
Charge (C) = Current (A) × Time (s)
Charge = 4.50 A × 8000 s = 36,000 C
Next, we will use Faraday's laws to convert the charge into moles of electrons:
Moles of electrons = Charge (C) / Faraday's constant
Moles of electrons = 36,000 C / 96,485 C/mol = 0.373 mol
Since the balanced equation shows that 1 mole of CrO4{2-} is reduced to 1 mole of Cr(OH)4-, we can conclude that the number of moles of Cr(OH)4- formed is also 0.373 mol.
Now, to find the mass of Cr(OH)4-, we need to multiply the number of moles of Cr(OH)4- by its molar mass.
The molar mass of Cr(OH)4- can be calculated by adding up the atomic masses of chromium (Cr), oxygen (O), and hydrogen (H).
Cr(OH)4-:
Cr: 1 atom × (atomic mass of Cr)
O: 4 atoms × (atomic mass of O)
H: 4 atoms × (atomic mass of H)
Using the atomic masses from the periodic table:
Cr: 51.996 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of Cr(OH)4- = (1 × 51.996 g/mol) + (4 × 16.00 g/mol) + (4 × 1.01 g/mol) = 169.036 g/mol
Now, we can calculate the mass of Cr(OH)4- using the formula:
Mass = Moles × Molar mass
Mass = 0.373 mol × 169.036 g/mol ≈ 63.051 g
Therefore, approximately 63.051 grams of Cr(OH)4- is formed by the passage of 4.50 A for a period of 8000 seconds.