We used a Cu wire and 0.5 M HC2H3O2 in 0.5 M Na2SO4
In the electrolysis cell we shall study, the reduction reactionof interest will occur in a slightly acidic medium; hydrogen gaswill be produced by the reduction of hydrogen ion: 2H+ (aq) + 2e--> H2 (g)
Known:
Number of moles of hydrogen gas produced in eachtrial: Trial 1: 0.00197 moles; Trial 2: 0.001837moles
Moles of electrons (faradays) used to produce the hydogen gas ineach trial: Trial 1: 0.00394 moles e and Trial 2: 0.003674 molese
Possible reaction for the anode: Cu(s)--> Cu2+ (aq) + 2e Ered= -0.34
and cathode: So42- (aq) + 4H+ (aq) + 2e ---> H2SO3 (aq) + H2o(l) Ered= 0.20
Calculate the following:
1. Knowing the charge of the cations that are produced when themetal anode is oxidized, calculate the moles of metal that wereoxidized in each trial
2. From the loss of mass of the anode and the moles of anodethat were oxidized during the electrolysis, calculate the molarmass of the metal in each trial
3. Compare your molar mass result to the actual value anddiscuss the main sources of error
4. What is being reduced and what is being oxidized? Write thereduction and oxidation half reactions and the overall reaction
5. If 0.5 amps of current flowed for 10 minutes, what volume ofhydrogen gas would be produced at STP (standard temperature andpressure = 0 degrees celcius, 1 atm). Charge= current X time inseconds
