To determine the loss or gain in weight of the electrodes when a current passes through two different electrolytic cells, we can use Faraday's laws of electrolysis. These laws relate the amount of substance transformed at an electrode to the quantity of electricity that passes through the electrolyte. Let's break this down step by step.
Understanding the Electrolytic Cells
In your scenario, we have two cells:
- Cell 1: Copper sulfate (CuSO4) solution with a copper (Cu) electrode.
- Cell 2: Silver nitrate (AgNO3) solution with a silver (Ag) electrode.
Calculating the Quantity of Electricity
First, we need to find out how much charge is required to produce 40 cc of oxygen gas at normal temperature and pressure (NTP). At NTP, 1 mole of gas occupies 22.4 liters, which is equivalent to 22,400 cc. Therefore, 40 cc of oxygen corresponds to:
Number of moles of O2:
\[
\text{Moles of O2} = \frac{40 \text{ cc}}{22400 \text{ cc/mole}} = 0.00179 \text{ moles}
\]
Since the electrolysis of water produces oxygen gas at the anode, we can use the stoichiometry of the reaction. The reaction for the production of oxygen is:
2 H2O → O2 + 4 H+ + 4 e-
This means that to produce 1 mole of O2, 4 moles of electrons are required. Therefore, for 0.00179 moles of O2, the moles of electrons needed are:
Moles of electrons:
\[
\text{Moles of e-} = 0.00179 \times 4 = 0.00716 \text{ moles}
\]
Now, using Faraday's constant (approximately 96500 C/mol), we can find the total charge (Q) required:
Total charge (Q):
\[
Q = \text{Moles of e-} \times \text{Faraday's constant} = 0.00716 \times 96500 \approx 691.4 \text{ C}
\]
Electrode Reactions and Weight Changes
Next, we analyze the reactions at each electrode:
At the Copper Electrode (Cu)
The reduction reaction at the cathode (where Cu2+ ions gain electrons) is:
Cu2+ + 2 e- → Cu
From this reaction, we see that 2 moles of electrons deposit 1 mole of copper. The molar mass of copper is approximately 63.5 g/mol. Therefore, the weight gain at the copper electrode can be calculated as follows:
Weight gain of Cu:
\[
\text{Weight gain} = \left( \frac{63.5 \text{ g}}{2} \right) \times \text{Moles of e-} = \left( \frac{63.5}{2} \right) \times 0.00716 \approx 0.227 \text{ g}
\]
At the Silver Electrode (Ag)
The reduction reaction at the cathode (where Ag+ ions gain electrons) is:
Ag+ + e- → Ag
Here, 1 mole of silver is deposited for every mole of electrons. The molar mass of silver is approximately 107.9 g/mol. Thus, the weight gain at the silver electrode is:
Weight gain of Ag:
\[
\text{Weight gain} = 107.9 \text{ g/mol} \times 0.00716 \text{ moles} \approx 0.771 \text{ g}
\]
Summary of Weight Changes
In summary, during the time it takes to collect 40 cc of oxygen gas:
- The copper electrode gains approximately 0.227 grams.
- The silver electrode gains approximately 0.771 grams.
This analysis illustrates how electrolysis can lead to measurable changes in the mass of electrodes based on the amount of current passed and the nature of the electrolyte solutions involved. If you have any further questions or need clarification on any part of this process, feel free to ask!
