when zinc in refined by electrolysis, the following reaction at the cathode is:
zn2(aq)+ +2e- ---->zn(s)
a competing reaction which lowers the yield is the formation of hydrogen gas
2H+ (aq) + 2e- ------>H2(g)
If 91.5 % of the current flowing results in zinc being deposited, while 8.5% produces hydrogen gas, how many liters of H2 (g), measured at STP, form per kg of zinc?
2007-12-04 05:14:09 · 3 answers · asked by skoolgrl217 2 in Science & Mathematics ➔ Chemistry
Because reduction of both Zn++ and H+ require the same 2e-, the same number of faradays and moles are gotten for each.
Atomic weights: Zn=65.4 H=1 H2=2
1kgZn x 1000gZn/1kgZn x 1molZn/65.4gZn x 8.5molH2/91.5molZn x 22.4LH2/1molH2 = 31.8 L H2 to three significant figures.
2007-12-04 05:26:32 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
The way I would approach this problem is by trying to figure out how many moles of H2 is produced.
If a kilogram of Zinc is deposited, how many moles of Zinc is that? 1,000/65.39 = 15.29 moles
If 15.29 moles is 91.5% of the moles produced of both H2 and Zn, then the total moles produced is 15.29/0.915 = 16.71 moles
The moles of H2 produced is the difference between the total produced and Zn produced: 16.71 - 15.29 = 1.42 moles of H2
The molar volume at STP is 22.414 liters/mole so we have produced 22.414 x 1.42 = 31.9 liters
2007-12-04 13:29:56 · answer #2 · answered by knottysailorboy 2 · 0⤊ 0⤋
wow that is a tough question....but i think you should use the formula no. of moles=volume of gas/rtp or stp depending on ur situation
2007-12-04 13:19:25 · answer #3 · answered by lp342 4 · 0⤊ 1⤋