A current of 15.6 A is passed through an electrolysis cell containing molten CaC
ID: 1045319 • Letter: A
Question
A current of 15.6A is passed through an electrolysis cell containing molten CaCl2 for 38.8minutes.
(a) Predict the products of the electrolysis and write the reactions occurring at the anode and cathode. Write electrons as e-. Use smallest integer coefficients possible and omit states. If a box is not needed, leave it blank.
Cathode reaction:
Anode reaction:
(b) Calculate the grams of metal or liters of gas formed at the cathode and anode (assume gases are at 298 K and 1.00 atm).
Formed at cathode: _______grams or liters
Formed at anode: _______grams or liters
Explanation / Answer
a. CaCl2 (aq) ------------------> Ca^2+ (aq) + 2Cl^- (aq)
at cathode reaction
Ca^2+ (aq) + 2e^- --------> Ca(s)
at anode reaction
2Cl^- (aq) ----------> Cl2 (g)+ 2e^-
b.
Ca^2+ (aq) + 2e^- --------> Ca(s)
W = MCt/ZF
Z =2
F = 96500c
M = 40g/mole
C = 15.6amp
t = 38.8 min = 38.8*60sec = 2328sec
W = 40*15.6*2328/2*96500 = 7.53g of Ca
2Cl^- (aq) ----------> Cl2 (g)+ 2e^-
Z =2
M = 71g/mole
W = MCt/ZF
W/M = Ct/ZF
n = Ct/ZF
= 15.6*2328/2*96500 = 0.188moles of Cl2
P = 1atm
T = 298K
PV = nRT
V = nRT/P
= 0.188*0.0821*298/1 = 4.6L of Cl2
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