Electrochemical calculations
YOU SHOULD BE ABLE TO:
1. Define the Faraday constant. 2. Perform calculations involving the mole. 3. Calculate the masses and volumes of substances liberated during electrolysis.
The Faraday (F) is the quantity of electricity needed to remove one mole of electrons from the cathode during electrolysis, or to deposit one mole of electrons on the anode during electrolysis.
The Faraday constant is the amount of electric charge carried by one mole of electrons, that is 96,500C.
During electrolysis, electrons are taken from the cathode by positive ions called cations.
E.g., 2H+(aq)+ 2e =H (g) Cu2+(aq)+ 2e = 2 Cu(s) The ions are said to be discharged. Electrons are deposited on the anode (+ve) by negative ions called anions. E.g., 2Cl- (aq)= Cl (g) + 2e 4OH- = 2H O+ O 2 2 2 + 4e
Electrons may also leave from the anode if the anode dissolves. E.g., Cu = Cu2+ + 2e
One Faraday = 96,500 Coulombs, i.e., 1F = 96,500C.
The coulomb is the unit of electrical charge and is 1 ampere flowing for 1 second.
Coulombs = amps x sec, quantity of electricity = current x time (Q= I x t)
The Faraday may also be regarded as the charge on 1 mole of electrons.
Thus F = Le; where L = Avogadro’s number; e = the charge on one electron.
EXAMPLE 1
What mass of copper is deposited on passing a current of 0.5A for 3,860 seconds? First, determine the electrode reaction. At the cathode: Cu2+ + 2e = Cu (s) From the equation, 2mol of electrons are needed to deposit 1 mole of Cu. Therefore, 2F are needed = 2x96,500 = 193, 000C 193,000C deposits 64g Cu (1mole)
Next, determine the actual amount of electrical charge flowing through the cell. Q = Ixt = 0.5x3860 = 1,930C If 193,000C produces 64g Cu Then 1,930C will produce X g Cu Solve for X X = (1,930x64)/ 193,000 = 0.64g Cu Mass of copper deposited = 0.64g Cu.
EXAMPLE 2
What mass of sodium is deposited by a current of 2A flowing for 9,650 seconds? Equation: Na+ +e = Na (l) 1 mole of electron requires 1F = 96,500C Therefore, 96,500C deposits 23g Na (1mole)
Quantity of electricity = I*t = 2*9650 = 19,300C If 96,500C deposits 23g Na Then 19,300C deposits X g Na X = (19,300x23)/ 96,500 = 4.6g Na Mass of sodium deposited = 4.6g Na
EXAMPLE 3
What volumes of (a) H2(b) O2 would be liberated at RTP when 2F of electricity is passed through dilute sulphuric acid? Equations: Cathode: 4H+ + 4e- = 2H2 Anode: 4OH- = 2H2O + O2 + 4eCalculate
the volume of H , for example, 2H+ 2 + 2e = H2 1 mole of hydrogen gas is liberated by 2 moles of electrons = 2F 1 mole of a gas at RTP = 24dm3
At the anode, oxygen is given off. From the equation, 1 mole of oxygen requires 4mol of electrons to be liberated. 4F would liberate 24dm3 of oxygen Since only 2F of electricity was passed, then the volume of oxygen produced would be halved. Volume of oxygen = 24/2 = 12dm3.
Volume = 24dm3 H and 12dm3 O 2 2