Electrochemistry Question: Calculation

[readerperson]

Hi there, this might be a rather elementary topic but here goes...

1) An electrolytic circuit using graphite electrodes and sulfuric acid as the electrolyte is set up.

One of the gases (A) is more than twice the volume of the other (B). Volume of B is taken as 10 cm3. It is found that vol. of A is 2.5 cm3 greater than what is expected. With appropriate equations, suggest why this is so.

I'm really not sure on this one. Did the hydrogen react with another substance?

2) 2 voltaic cells: Cell A using Mg and Cu and Cell B using Zn and Cu is set up. Electrolyte used is dilute sulfuric acid.

(a) Assuming rate of current is same for both cells, which cell will have a greater loss in mass at one of the electrodes and why?

My Attempt: Cell B ( I don't know why. )

(b) Calculate the loss in mass of the electrote in cell B when the loss in mass in cell A is 1.2g.

My attempt: No. of moles of Mg lost = 1.2/24.3 = 0.4938271
Hence, mass of Zngained is: 0.4938271 x 65.4 = 32.29629

My answers are quite feeble attempts and they're most probably wrong. I would like someone to explain. Thanks in advance!

 

[nate808]

Hi there,

I can help you understand the concepts behind these questions.

1) In an electrolytic circuit, the flow of electricity causes a chemical reaction to occur at the electrodes. In this case, the graphite electrodes are reacting with the sulfuric acid to produce gases. The gas with a volume of 10 cm3 is likely hydrogen gas (H2), while the gas with a volume of 2.5 cm3 is likely oxygen gas (O2). This is because the ratio of hydrogen to oxygen in water (which is produced when sulfuric acid is electrolyzed) is 2:1. However, the excess volume of oxygen gas could be due to impurities in the sulfuric acid or a possible side reaction occurring at the electrodes.

2) In voltaic cells, the flow of electrons between the electrodes causes a redox reaction to occur. In Cell A, the magnesium (Mg) is oxidized and the copper (Cu) is reduced, while in Cell B, the zinc (Zn) is oxidized and the copper (Cu) is reduced. The rate of current is the same for both cells, but the rate of the redox reaction may be different. This means that the rate at which the electrodes are losing or gaining mass may also be different. Since zinc has a higher reactivity than magnesium, it is likely that Cell B will have a greater loss in mass at one of the electrodes.

(a) The loss in mass at one of the electrodes in Cell B will be greater because the redox reaction is happening at a faster rate.

(b) Your attempt is correct, but you forgot to take into account the molar mass of zinc (65.4 g/mol). The correct calculation would be: 0.4938271 x 65.4 = 32.29629 g. This means that for every 1.2 g of magnesium lost in Cell A, 32.29629 g of zinc will be gained in Cell B.

I hope this helps to clarify the concepts behind these questions. Let me know if you have any further questions. Keep up the curiosity and good work!