Exp. Electrochemistry Advanced Study Questions Name A solution of 500.0 mL of 0.

Question

Exp. Electrochemistry Advanced Study Questions Name 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'ad)+ 2e- Hz (a) The hydrogen gas collected in this reaction occupied a volume of 98.00 mL at 22.0 °C and 738.5mm Hg. 1. How many moles of hydrogen gas were produced? mol Ha 2. How many moles of Ht ions were reduced? , mol H. 3. What was the final concentration of Ht ions in the solution after the reaction stopped? 4. How long, in minutes, did it take to produce this gas sample if a current of 0.50A was used _ min

Explanation / Answer

1) Ideal gas equation is

PV = nRT

n= PV/RT

= 0.9748atm × 0.098L /0.082057(L atm/mol K)×295.15K

= 0.003944

so, 0.003944 mole of H2 is produced

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

stoichiometrically, 1mole of H2 is formed by 2mole of H+ ion

0.003944mole of H2 is actually produced

So, 0.007888mole of H+ is reduced

3) Initial mole of H+ = (0.500mol/1000ml)×500ml = 0.250mol

No of mole of H+ reduced = 0.007888

remaining mole of H+ = 0.250 - 0.007888 = 0.24211mol

Final concentration = (0.24211mol/500ml)×1000ml = 0.4842M

4) 2mole electron is required to produce 1mole of gas

No of mole of H2 produced = 0.003944mol

No of mole of electron consumed = 2× 0.003944 = 0.007888 mol

Coloumbs of 0.007888 mole of electron = (96485Colombs/1mol)×0.007888Colomb= 761.07Coloumbs

0.50A = 0.50Coloumbs per seconds

Time consumed = (1seconds/0.50Coloumbs)×761.07Coloumbs = 1522.1seconds = 25.37minutes

So, the answer is 25.37minutes

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