96,485 coulombs will deposit 27/3 = 9 grams of Al. This problem has 9.14 g Al deposited; therefore, 96,485 C x 9.14/9 = 97,986 C used.
96,485 C will deposit 2.02 grams H2 gas @ STP so 97,986 C will release
2.02 x 97,986/96,485 = 2.05 grams H2 @ STP.
Then use PV = nRT to determine the volume at the conditions listed. For n use n = grams H2/molar mass H2 = 2.05/2.02 = ?
Don't forget that temperature must be convert to kelvin. K =273 + C
Post your work if you get stuck.
A certain quantity of electric charge deposited 9.14 g of Al at the cathode during electrolysis of a solution of Al^+3(aq). What volume of H2(g) will be discharged at 30°C and 770mmHg if the same quantity of electricity is used in the reduction of H^+(aq) at a cathode
2 answers
Thanks Sir 👍
(Workings)
From....n =2.05/2.02=1.01
V = nRT/P
= (1.01×8.314×303)÷(770)
V = 3.03cm^3
(Workings)
From....n =2.05/2.02=1.01
V = nRT/P
= (1.01×8.314×303)÷(770)
V = 3.03cm^3