11someone please help [2]
1)The water is electrolysed in a cell, hydrogen is liberated at one electrode and oxygen simultaneously liberated at the other. In a particular experiment hydrogen and oxygen so produced were collected together and the total volume measured at 16.8 ml at NTP. How many coulombs were passed through the cell in the experiment?
2)50 ml of hydrogen gas was collected over at 296 K, 740 mmHg barometric pressure. H2 was produced by the electrolysis of water. The voltage was constant at 2.1 volts, the current averaged 0.5 ampere for 12 minutes and 20 seconds. Calculate Avogadro constant.
3)50 ml of a 0.1 M CuSO4 solution was electrolysed for 12 minutes at a constant 0.06 ampere. If Cu is produced at one electrode and O2 at the other, what will be the pH of the final solution?
for HSO4---------> H+ + SO4-, Kdiss = 0.013
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8 Answers
1first one is easy.....
keep in mind that inert electrodes are used
Passing 1F of electricity,
At Anode
4OH- = O2 + H2O + 4e-
here 'n' factor of O2 is 4....
gm eq of O2 produced = 1
hence mol of O2 = 1/4 => volume=5.6 L
Similarly mol of H2 = 1/2 => volume=11.2 L
total volume of gases = 16.8 L
Now 16.8 L ≡ 96500 C
or 16.8 mL ≡ 96.5 C
112)50 ml of hydrogen gas was collected over at 296 K, 740 mmHg barometric pressure. H2 was produced by the electrolysis of water. The voltage was constant at 2.1 volts, the current averaged 0.5 ampere for 12 minutes and 20 seconds. Calculate Avogadro constant.
2). Using above reaction every molecule of H2 produced requires 2 electrons. From the
problem using PV = nRT you can calculate the moles of H2. (Moles = number of molecules /
NA, but we dont know NA and number of molecules).
So lets calculate number of molecules = 0.5 x number of electrons transferred.
0.5 A current for 12 x 60 + 20 = 740 sec ==> 0.5 x 740 = 370 C.
Number of electrons = 370 / Charge on elctron.
Now calculate Av. constant
233] 2H2O → O2 + 4e- + 4H+
current and all is given , so u can find how much new H+ is produced. add dis to the initial conc of H+ , and u get ur pH