A solution of 500.0 mL of 0.500 M strong acid HA was used as part of an electrol

Question

A solution of 500.0 mL of 0.500 M strong acid HA was used as part of an electrolytic cell.
The reduction half-reaction that took place was:
2 H+(aq) + 2e --> H2 (g)
The hydrogen gas collected in this reaction occupied a volume of 98.00 mL at 22.0 oC
and 738.5mm Hg.
1. How many moles of hydrogen gas were produced?
________ mol H2
2. How many moles of H+ ions were reduced?
________ mol H+
3. What was the final concentration of H+ ions in the solution after the reaction stopped?
________ M
4. How long, in minutes, did it take to produce this gas sample if a current of 0.50A was
used.
________ min

Explanation / Answer

Ans 1 :

The number of moles can be calculated by using the formula , pV = nRT

738.5 mm Hg = 0.97171 atm

22 degree C = 295.15 K

So putting all the values we get :

0.97171 x 0.098 = n x 0.0821 x 295.15

n = 0.0039 moles

Ans 2 :

2 moles of H= reduce to form 1 mole of H2

So number of moles of H+ will be = 2 x 0.0039

= 0.0078 moles

Ans 3 :

Molarity = no. of moles of H+ / Volume of solution in L

Total number of moles of H+ = 0.500 x 0.500 = 0.25 moles

Moles of H+ utilised = 0.25 - 0.0078 = 0.2422

Molarity = 0.2422 / 0.500

= 0.4844 M

Ans 4 :

Number of moles of H+ = number of moles of electrons = 0.0078

Charge = 0.0078 x 96487 = 752.6 Coloumb

Q = I. t

752.6 = 0.50 x t

t = 1505 seconds

or 25 minutes

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