A solution of Co(NO₃)₂ is electrolyzed using a 15.0A current for 1.00 hour and Co(s) is produced at the cathode. Calculate the moles of electrons that was used to electrolyze the solution? Calculate the moles and mass of Co(s) produced at the cathode?

1. 0.56 mole of electron.

2. 0.28 mole of Co .

3. 16.52g of Co.

Explanation:

The following data were obtained from the question:

Current (I) = 15A

Time (t) = 1 hr = 1 x 60 x 60 = 3600secs

Next, we shall determine the quantity of electricity, Q flowing in electrolyte. This is illustrated below:

Quantity of electricity (Q) = current (I) x Time

Q = It

Q = 15 x 3600

Q = 54000C.

1. Determination of the moles of electrons used to electrolyzed the solution. This is illustrated below:

96500C = 1 mole of electron

Therefore, 54000C = 54000/96500 = 0.56 mole of electron.

Therefore, 0.56 mole of electron was used to electrolyze the solution.

2. Determination of the number of mole of Co produced at the cathode. This is illustrated below:

At the cathode:

Co2+ + 2e —> Co

From the balanced equation above,

2 moles of electron produce 1 mole of Co.

Therefore, 0.56 mole of electron will produce = 0.56/2 = 0.28 mole of Co.

Therefore, 0.28 mole of Co is produced at the cathode.

3. Determination of the mass of Co produced at the cathode. This can be achieved by doing the following:

Molar mass of Co = 59g/mol

Number of mole of Co = 0.28 mole

Mass of Co =..?

Mass = mole x molar mass

Mass of Co = 0.28 x 59

Mass of Co = 16.52g

Therefore, 16.52g of Co is produced at the cathode.