Question

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?

Answer 1

1)

a.

E^o_cell = E^o_cathode - E^o_anode

= 0.15 V - (-1.25 V)

= + 1.40 V.

Free energy change (G^o ) = - nFE^o_cell

Number of electron transferred (n) = 2.

F = 96500 C/mol

G^o = 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.

E^o_cell = E^o_cathode - E^o_anode = 0.15 V - (-0.76 V ) = 0.91 V.

Number of electron transferred (n) = 2

G^o = - 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.