An antique automobile bumper is to be chrome plated. The bumper, which is dipped into an acidic Cr₂O₇²^- solution, serves as a cathode of an electrolytic cell. The atomic mass of Cr is 51.996; 1 faraday = 96,485 coulombs.
-If oxidation of H₂O occurs at the anode, how many moles of oxygen gas will evolve for every 2.98 * 10² g of Cr(s) deposited?

Question

Quizplus · Accepted Answer

The balanced half-reactions are:

Cathode: Cr3+ + 3e- → Cr(s)

Anode: 4HSO4- → 2S2O8^2- + 4H+ + 3O2 + 4e-

The overall reaction is:

Cr3+ + 4HSO4- → Cr(s) + 2S2O8^2- + 4H+ + 3O2

We need to find the moles of oxygen gas evolved per 2.98 × 10^10 g of Cr(s) deposited.

First, we need to calculate the amount of Cr(s) that is deposited. From the balanced equation, 3 moles of electrons are required to deposit 1 mole of Cr. The charge on one mole of electrons is:

1 mol e- × 96,485 C/mol e- = 96,485 C

The charge required to deposit 1 mole of Cr can be calculated as follows:

1 mol Cr × (3 mol e- / 1 mol Cr) × (96,485 C/mol e-) = 2.91 × 10^5 C

The total charge required to deposit 2.98 × 10^10 g of Cr(s) can be calculated as follows:

2.98 × 10^10 g Cr × (1 mol Cr / 51.996 g) × (2.91 × 10^5 C/mol Cr) = 1.75 × 10^11 C

The amount of oxygen gas evolved can be calculated from the stoichiometry of the reaction:

3 mol O2 / 4 mol e- = 0.75 mol O2 / mol e-

The total moles of oxygen gas evolved can be calculated as follows:

1.75 × 10^11 C × (1 mol e- / 96,485 C) × (0.75 mol O2 / mol e-) = 1.45 × 10^6 mol O2

The moles of oxygen gas evolved per 2.98 × 10^10 g of Cr(s) can be calculated as follows:

1.45 × 10^6 mol O2 / (2.98 × 10^10 g Cr × (1 mol Cr / 51.996 g)) = 8.60 mol O2 / 2.98 × 10^10 g of Cr(s)

Therefore, the answer is B: 8.60 mol.

An Antique Automobile Bumper Is to Be Chrome Plated