Question

Could someone please explain how to do this in detail please? If an aqueous solution of HNO₃ is electrolyzed for 24.00 min at a steady current of 1.09 A, what volume of H₂ (g) at 25.0^oC and 0.99 atm will be collected at the cathode?

Answer 1

2H+ + 2e- ------> H2

1 mol of H2 requires 2 mol of electron

1 mol of electron = 96485 C

So,1 mol of H requires 192970 C

let us calculate the charge passed:

t = 24.0 min = 24.0*60 s = 1440 s

time, t = 1440s

Q = I*t

= 1.09 A * 1440 s

= 1569.6 C

mol of H2 released = 1569.6/192970 = 0.00813 mol

Given:

P = 0.99 atm

n = 0.00813 mol

T = 25.0 oC

= (25.0+273) K

= 298 K

use:

P * V = n*R*T

0.99 atm * V = 0.00813 mol* 0.0821 atm.L/mol.K * 298 K

V = 0.201 L

V = 201 mL

Answer: 201 mL