Electrolysis is carried out for 2.00 h in the following cell. The platinum cathode which has a mass of 25.0782 g, weighs 25.8639g after the electrolysis. The platinum anode weighs the same before and after the electrolysis.
there is H2SO4(aq) on the anode side and AgNO3(aq) on the cathode side.
a) write equations for the half reactions that occur at the two electrodes.
b)what must have been the current used assuming a constant current throughout?
c)a gas is collected at the anode. what is it? how many moles of it are present?
a. Ag^+(aq) + e ==> Ag(s)
4OH^- ==> 4e + 2H2O + O2
b. 25.8639
..-25.0782
mass Ag metal deposited = ?
96,485 coulombs will deposit 107.868 g Ag so
C used = 96,485 x (?g Ag/107.868) = ?coulombs.
C = amperes x seconds.
Substitute and solve for amperes.
c. ?C from above = 96,485 x ?gO2/8
Substitute and solve for ?g O2.
a) The half reactions that occur at the two electrodes can be determined based on the electrolyte solutions present.
At the cathode:
AgNO3(aq) + e- → Ag(s)
At the anode:
2H2O(l) → O2(g) + 4H+(aq) + 4e-
b) To determine the current used, we need to use the concept of Faraday's Law of Electrolysis. The formula for Faraday's Law is given by:
Q = nF
where Q is the total electric charge (current × time), n is the number of moles of electrons transferred, and F is the Faraday constant.
Since the platinum cathode gained mass, it means that silver ions from AgNO3(aq) were reduced to Ag(s) at the cathode. The number of moles of silver formed can be calculated from the change in mass of the cathode.
Δm = mass of cathode after electrolysis - mass of cathode before electrolysis
n_Ag = Δm / M_Ag
where M_Ag is the molar mass of silver.
Similarly, since the platinum anode did not change in weight, no reaction occurred at the anode. Thus, no moles of electrons were transferred from the anode side. This means that the same number of moles of electrons that were consumed at the cathode must have been produced by the current source.
Therefore, the current used is given by:
Current = n_Ag × F / time
c) The gas collected at the anode is oxygen (O2) gas, which is produced from the oxidation of water. The number of moles of oxygen gas can be determined using the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
In this case, the volume and pressure of the gas are not provided, so we cannot directly calculate the number of moles of oxygen present. However, if we assume standard temperature and pressure (STP), we can determine the number of moles of oxygen gas produced from the given reaction:
2H2O(l) → O2(g) + 4H+(aq) + 4e-
Since two moles of water produce 1 mole of oxygen gas, the number of moles of oxygen gas can be calculated as:
n_O2 = 0.5 * n_Ag (since each mole of Ag formed requires two moles of electrons)
Substituting the value of n_Ag calculated earlier, you can determine the number of moles of oxygen gas produced.