In: Chemistry
1.) The cell reactions occurring in a battery are given by:
Cathode : 2MnO2 (s) + H2O (l) + 2e-→ Mn2O3 (s) + 2HO- (aq) E 0 red = +0.15 V
Anode : Zn (s) + 2HO- (aq) → Zn(OH)2 (s) + 2e- E 0 red = -1.25 V
a) What is the overall cell potential? What is the free energy change for this process?
b) During the discharge of the battery, 2.00 g of Mn2O3 is produced at the cathode. How many grams of Zn were consumed?
c) Consider this half reaction: Cd(OH)2 + 2e- → Cd(s) + 2HO- (aq) E°red = -0.76V What is the overall cell potential and free energy change of the battery described above if the Zn anode is replaced by Cd?
2.) Ca(s) can be obtained by electrolysis of molten CaCl2. What amperage is needed to produce 10.00 g of Ca(s) in a period of 1h assuming that the process is 100% efficient?
1)
a.
Eocell = Eocathode - Eoanode
= 0.15 V - (-1.25 V)
= + 1.40 V.
Free energy change (Go ) = - nFEocell
Number of electron transferred (n) = 2.
F = 96500 C/mol
Go = 2× 96500 × 1.40 = - 270200 J/mole = - 270.2 KJ/mole.
b.
Moles of Mn2O3 = ( mass/molar mass)
= [ 2.00 g/158 (g/mol)]
= 0.0126
The net reaction is
Zn + 2MnO2 + H2O Zn(OH)2 + Mn2O3
So,
Moles of Mn2O3 is produced = moles of Zn is consumed
= 0.0126.
Then, moles of Zn consumed = 0.0126
Mass of Zn = moles × molar mass of Zn
= 0.0126 (mol) × 65.38 (g/mol)
= 0.827 g.
C)
Now, if Cd is anode.
Eocell = Eocathode - Eoanode = 0.15 V - (-0.76 V ) = 0.91 V.
Number of electron transferred (n) = 2
Go = - 2× 96500 × 0.91 = - 175630 J/mol = - 175.63 KJ/mol.
2.)
By Faraday's law of electrolysis
W =
Given, mass of Ca (W) = 10.00 g.
Gram equivalent of Ca (E) = 20.00 g.
i = amperage
t = 1 h = 3600 sec.
Now,
10.00 =
Or, 20.00× i × 3600 = 96500× 10.00
Or, i = (965000/20.00×3600)
Or, i = 13.40 amp.