1)How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesiu
ID: 1019600 • Letter: 1
Question
1)How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten MgCl2 with an applied emf of 5.00 V?
a)0.0201
b)0.0496
c)12.4
d)49.6
e)24.8
a)0.56
b)0.50
0.44
0.40
0.52
A)2H+ + 2e H2
b)Sn Sn2+ + 2e
c)2H+ H2 + 2e
d)Sn + 2e Sn2+
e)Sn + 2e- H2
3) The standard emf for the cell using the overall cell reaction below is +0.48 V:
Zn (s) + Ni2+ (aq) Zn2+ (aq) + Ni (s)
The emf generated by the cell when [Ni2+] = 0.100 M and [Zn2+] = 2.25 M is
a)0.56
b)0.50
c)0.44
d)0.40
e)0.52
Explanation / Answer
According to Faraday first law,
m = Z * i * t
4.5 * 10^3 = 0.45341g.amp/hr *I * 1 hr
I = 9924.79 amp
Power = I * V = 9924.79 * 5
Power = 49.623 kW-hour
Answer ; D
2H+ (s) ------> H2(g) is the reduction reaction which occurs at cathode half cell
Sn (s) ------> Sn+2 (aq) is Oxidation reaction which occurs at Anode half cell
Answer : A
Zn (s) + Ni2+ (aq) -----> Zn2+ (aq) + Ni (s)
E cell = E0 cell -0.0591/nlog[Zn+2]/[Ni+2]
Ecell = 0.48 - 0.0591/2 log(2.25/0.1)
Ecell = 0.44 V
Answer : C
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