redox reactions - Faradays laws and calculation in electrochemistry- exercises with two cells-solution 1

1) A spoon was placed in a silver plating cell for 40 minutes with a current of 30.3A. Calculate the mass of silver deposited on the spoon.

Step 1 Calculate the charge passed through the cell during the 15 minute period.

Q = I X t
Q = 30.3 X (40 X 60)s = 72,720C
(time must be expressed in seconds)

Step 2 Calculate the mole of electrons passed through the circuit.

mole = 72,720C/96,500C/mol = 0.754 mole.

Step 3 Calculate the mass of silver

According to the equation above 0.754 mole of electrons will produce 0.754 mole of silver.
The mass of silver is therefore 0.754 X 107.9 = 81.31grams.

Hide the solution

A copper plating cell operates for 20 minutes with a steady current of 25.0A. What mass of copper is deposited?

Step 1 Calculate the charge passed through the cell during the 20 minute period.

Q = I X t
Q = 25.0 X (20 X 60)s = 30,000C
(time must be expressed in seconds)

Step 2 Calculate the mole of electrons passed through the circuit.

mole = 30,000C/96,500C/mol = 0.311 mole.

Step 3 Calculate the mass of copper

According to the equation above 0.311 mole of electrons will produce 0.311/2 mole of copper
The mass of copper is therefore 0.155 X 63.54 = 9.88grams.

Hide the solution

An iron window frame is placed in an aluminium plating cell for 53 minutes with a steady current of 34.5A. What mass of aluminium is deposited?

Step 1 Calculate the charge passed through the cell during the 53 minute period.

Q = I X t
Q = 34.5 X (53 X 60)s = 109,710C
(time must be expressed in seconds)

Step 2 Calculate the mole of electrons passed through the circuit.

mole = 109,710C/96,500C/mol = 1.14 mole.

Step 3 Calculate the mass of aluminium

According to the equation above 1.14 mole of electrons will produce 0.1.14/3 mole of aluminium.
The mass of aluminium is therefore 0.379 X 27 = 10.23grams.

Hide the solution

60g of silver must be evenly deposited on a serving tray. The iron serving tray is placed in a silver plating cell with a steady current of 45.5A. How long should the tray be left in the cell for?

Step 1 Calculate the mole of silver deposited

mole =m/M = 60/107 = 0.561mole

Step 2 Calculate the mole of electrons passed through the circuit.
According to the equation below one mole of silver needs one mole of eelctrons to deposit.

mole = 0.561mol of silver= 0.561 mol of electrons

Step 3. Calculate the charge of 0.561 mol of electrons

0.561 X 96,500 = 54,136.5

Step 4 Calculate the time needed.

Q =I X t
54,136.5 = 45.5 X t
54,136.5/45.5 = t = 1,190s = 19.8 minutes

Hide the solution

An extruded iron pipe is placed in a copper plating cell. A mass of 300 grams of copper must be deposited evenly along its length. The pipe is left in the cell for 1 hour, to what current should the operator set the dial?

Step 1 Calculate the mole of copper

300 / 63.5 = 4.72 mole

Step 2 Calculate the mole of electrons needed.

According to the equation above we need twice as many mole of electrons as mole of copper.
mole of electrons = 4.72 X 2 = 9.45mole

Step 3 Calculate the charge of 9.45mole of electrons

Q = mole of electrons X 96,500 = 9.45 X 96,500 = 911,811C

Step 4 Calculate the current needed.

911,811 = I X 60 X 60

I =911,811/3600 = 253.2A

Hide the solution

An iron serving tray has a surface area of 350 square centimetres. Silver is to be deposited to a depth of 4mm in an silver plating cell. If the current delivered is a steady 34.6A how long should the operator keep the tray in the cell for?
Density of silver =

Step 1 Calculate the mass of silver needed.

Density X volume = mass
The volume of silver deposited is 350 X 0.4 =140 cubic cm.
10.54 X 140 = 1475.6 grams

Step 2 Calculate the mole of silver

1475.6 /107.9 =13.68mole of silver

Step 3 Calculate the mole of electrons needed

13.68 mole of electrons is needed according to the equation above.

Step 4 Calculate the charge on the 13.68 mole of electrons

13.68 X 96.500 = 13,196,978C

Step 5 Calculate the time needed.

Q =I X t

13,196,978 = 34.6 X t

t = 13,196,978/34.6 = 38141s = 10.6 hours

Hide the solution

A teapot with a surface area of 0.05 square metres is placed in a silver plating cell for 34 minutes with a steady current of 1.2A. If the silver is deposited evenly how deep is the layer deposited on the surface of the teapot?
Density of silver =

Step 1 Calculate the charge delivered.

Q =I X t
Q = 1.2 X 34 X 60= 2448C

Step 2 Calculate the mole of electrons

2448/96500=0.0254mole

Step 3 Calculate the mole of silver needed

0.0245 mole of silver is needed according to the equation above.

Step 4 Calculate the mass silver

0.0245 X 107.9 = 2.64g

Step 5 Calculate the volume of silver that is required

volume =mass / density

volume = 2.64/10.54 =0.25 cubic cm

Step 6 Calculate the depth

Volume = area x depth
depth(cm) = 0.25 / 500 sq cm =0.0005cm

Hide the solution

An unknown metal which forms ions with a charge of 3+ () was deposited on a small key ring in an electrolytic cell. A steady current of 4.6A was used for 34 minutes and a mass of 1.815 grams was deposited on the key ring. Calculate the relative atomic mass of the element (X) and attempt to identify it.

Step 1 Calculate the charge delivered

Q = I X t
4.6 X 34 X 60 = 9384C

Step 2 Calculate the mole of electrons delivered

9384 /96500 =0.097mole

Step 3 Calculate the mole of metal "X" deposited according to the reaction below.

0.097 / 3 = 0.0324 mole of "X"

Step 4 Calculate the relative atomic mass of "X"

mole = mass/ rel.atomic mass

mole = 1.815 / M

M =1.815/0.0324

M =56

It is most likely iron

Hide the solution

A metal with a cation of charge 2+ was deposited in an electrolytic cell using a steady current of 200A for 55 minutes. If 136.79 grams was deposited identify the metal.

Step 1 Calculate the charge delivered

Q = I X t
200 X 55 X 60 = 660,000C

Step 2 Calculate the mole of electrons delivered

660,000 /96500 =6.84mole

Step 3 Calculate the mole of metal "X" deposited according to the reaction below.

6.84 / 2 = 3.42 mole of "X"

Step 4 Calculate the relative atomic mass of "X"

mole = mass/ rel.atomic mass

mole = 136.79 / M

M =136.79/3.42

M =40

It is most likely calcium

Hide the solution

Electrolysis exercises continued.

Two electrolytic cells are connected in series as shown on the right. A current of 2.00 A flows for 10.0 minutes.

a) What is the charge flowing through the cells?

b) Give the half equations for the reactions occuring at electrodes A to D when current is flowing.

c) What is the change in mass of electrode C after 10.0 minutes.

d) How does the reaction occuring at electrode B change after a while?

Quantity of charge passed through the cells in coulomb
=> Q = I x t (sec)
=> Q = 2.00 X 60 X 10.0 = 1200 C

At electrode A we have oxidation taking place.

We look for the reductants present and select the strongest.
Only one reductant exists at electrode A, namely H₂O.
2H₂O_(l) => O_2(g) + 4H⁺_(aq) + 4e

At electrode B we have reduction taking place.

We look for the oxidants present and select the strongest.
Two oxidants exist at electrode B, namely Al³⁺ and H₂O.

Now we select the strongest according to the electrochemical series on the right, which is H₂O.
2H₂O_(l) + 2e => H_2(g) + 2OH^-_(aq)

At electrode C we have oxidation taking place.

We look for the reductants present and select the strongest.
Two reductants exist at electrode C, namely Cu and H₂O

Now we select the strongest according to the electrochemical series on the right, which is Cu.
Zn_(s) =>Zn²⁺_(aq) + 2e

At electrode D we have reduction taking place.

We look for the oxidants present and select the strongest.
Two oxidants exist at electrode D, namely H⁺_(aq) and H₂O

Now we select the strongest according to the electrochemical series on the right, which is H⁺.
2H⁺_(aq) + 2e => H_2(g)

The reaction Zn_(s) => Zn²⁺_(aq) + 2e takes place at electrode C. Obviously we will lose zinc.

Step 1 Find the mol of electrons that flow through the circuit.

=> 1200 / 96,500 = 0.0124

Step 2 Calculate the mol of Zn that are used up.
=> 0.0124 / 2 = 0.0062

Step 3 Find the mass of zinc
=> 0.0062 X 65.4 = 0.405 grams lost

As the H⁺ decreases in concentration the next strongest oxidant will react. By this time we should have enough Zn²⁺ ions in solution to initiate the following reaction.

Zn²⁺_(aq) + 2e =>Zn_(s)

Home