If 1.70 liters of 0.670 M CuSO4 solution is electrolyzed by passing 4.70 amps through the solution for 3.00 hr using inert electrodes.
(a) As electrolysis proceeds, how does the acidity of the solution change?
(b) How many grams of Cu metal are formed at the cathode?
(c) How many liters of oxygen gas are collected at 29.5°C and 730. mm Hg?
(d) What is the [Cu2+] in the solution at the end of the electrolysis?
To answer these questions, we need to apply the principles of electrochemistry and stoichiometry. Let's go step by step:
(a) To determine how the acidity of the solution changes during electrolysis, we need to consider the reaction that occurs at the anode and cathode. In this case, the anode is the positive electrode, and the cathode is the negative electrode. Since we're using inert electrodes, there won't be any reactions happening at the electrodes. Instead, water will undergo electrolysis.
At the anode: 2H2O(l) -> O2(g) + 4H+(aq) + 4e-
At the cathode: Cu2+(aq) + 2e- -> Cu(s)
From these reactions, we can see that hydrogen ions (H+) are released at the anode, making the solution more acidic. Therefore, as electrolysis proceeds, the acidity of the solution increases.
(b) To determine the number of grams of Cu metal formed at the cathode, we need to calculate the moles of electrons transferred (n). We can use Faraday's law to do this:
n = (I * t) / F
Where I is the current in amperes, t is the time in seconds, and F is Faraday's constant (96500 C/mol).
In this case, I = 4.70 A and t = 3.00 hours = 3.00 * 60 * 60 seconds. Plugging these values into the equation gives:
n = (4.70 * 3.00 * 60 * 60) / 96500
Once we have the moles of electrons transferred, we can use the stoichiometry of the cathode reaction (Cu2+(aq) + 2e- -> Cu(s)) to find the moles of Cu metal formed.
Finally, using the molar mass of Cu, we can convert moles to grams.
(c) To determine the number of liters of oxygen gas collected, we can use the Ideal Gas Law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, convert the temperature from Celsius to Kelvin (T = 29.5 + 273.15). Then, use the given pressure (730 mm Hg) and convert it to atm (1 atm = 760 mm Hg).
Next, we need to find the moles of oxygen gas produced. We can use Faraday's law again, but this time for the anode reaction (2H2O(l) -> O2(g) + 4H+(aq) + 4e-). With this information, we can determine the volume of oxygen gas collected.
(d) To find the [Cu2+] in the solution at the end of the electrolysis, we need to consider the initial concentration and the amount of Cu2+ that has been reduced at the cathode. We can start by calculating the moles of Cu2+ reduced using the formula:
moles = (n * M) / (2 * F)
Where n is the moles of electrons transferred (calculated in part (b)), M is the molar mass of Cu, and F is Faraday's constant. Divide the moles by the total volume of the solution (1.70 L) to find the final concentration of Cu2+.
Now that we've discussed the steps and formulas involved, you can apply these calculations to find the answers to each question.