This was a lab on employing electrolysis to measure the amount of charge require

Question

This was a lab on employing electrolysis to measure the amount of charge required to reduce 1 mol of H+ ions according to:

2H+ (aq) 2e- -> H2 (g)

We used 500mL of 1 M H2SO4 solution.

a) What is the relationship between current and rate at which H2 produced? (Current was 0.71amp, time elapsed was 3min 20seconds, don't know if that helps or not!)

b) Would an increase in the concentration of sulfuric acid increase the rate of hydrogen production? Explain.

c) Why are different products obtained when the molten and aqueous NaCl are electrolyzed? Predict the products in each case.

d) What are the major sources of error, and how could they be minimized for electrolysis?

Any explanations are highly appreciated! Thank you very much (:

Explanation / Answer

hydrogen (-) 2H+(aq) + 2e-==> H2(g)

2 (2g) = 0.5 mol H2 gas (12 dm3) released per mol e-s

b)  Since water produces very few hydrogen ions, the addition of sulfuric acid would increase the rate of hydrogen production as sulfuric acid is 100% ionised in water and the concentration of avialable hydrogen ions will be much higher.

c) With molten NaCl , only Na+ and Cl- can be liberated so products are chlorine and sodium.
Na + + e- ---> Na
2Cl- - 2e- ----> Cl2
In aqueous solution there is a choice of ions as some H+and OH- from the water must be considered
As it is much easier to discharge H+ this happens at the cathode
2H+ + 2e- -----> H2
The chloride ion is still disharged at the anode in a normal salt solution.

d)  The cause of your voltage drops are the actual electrical resistances in the "circuits", the cells themselves". The cell potentials that you calculate are the "ideal" situation and you would get those if there was not some electrical resistance. But like every machine has some friction, every circuit has some resistance, and the affect of the resistance is to lower the potential difference, the voltage of the cell.

The internal resistance is not something you can get rid of. You can minimize it by using "clean" electrodes and keeping the distances between electrodes short.

You may also have an error in your voltmeter. You want to use a voltmeter with a high input impedance (resistance), ideally around 10 megohms. Otherwise, the voltmeter itself becomes part of the circuit and gives false readings. Avoid cheap voltmeters with lower input impedances.

hydrogen (-) 2H+(aq) + 2e-==> H2(g)

2 (2g) = 0.5 mol H2 gas (12 dm3) released per mol e-s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.