Question

A Zn-Cu battery is constructed as follows at 25^oC

Zn_(s) | Zn⁺²_(aq) (0.20 M) || Cu⁺²_(aq) (2.5 M) | Cu_(s)

The mass of each electrode is 200.0 gram and the volume of the electrolytes is 1.0 L

a) calculate the cell potential of the cell

b) calculate the mass of each electrode after 10.0 amp. of current has flowed for 10.0 hour Remember, the anode loses mass because of the oxidation reaction while the cathode gains mass due to the reduction process.

c) how long can this battery deliver a current of 10.0 amp before it goes dead?

Answer 1

a) calculate the cell potential of the cell

Ecell = E°cell - 0.0592/n * log(Q)

Q = [Zn+2]/[Cu+2] = 0.2/2.5 = 0.08

n = 2 electrons

Zn2+ + 2 e− ⇌ Zn(s) −0.7618

Cu2+ + 2 e− ⇌ Cu(s) +0.337

E°cell = Ered - Eoix = 0.337 - -0.7618 = 1.0988 V

Ecell = E°cell - 0.0592/n * log(Q)

Ecell =1.0988 - 0.0592/2* log(0.08) = 1.1312V

b) calculate the mass of each electrode after 10.0 amp. of current has flowed for 10.0 hour Remember, the anode loses mass because of the oxidation reaction while the cathode gains mass due to the reduction process.

t = 10 h = 10*3600 = 36000 seconds

total charge C = I*t = 10*36000 = 360000 C

96500 C/mol so..

mol of e- = 360000 /96500 = 3.7305 mol of e-

so...

mass lost by Zn --> mol*MW = 3.7305/2*65.38 = 121.9500 g

mass gained by Cu --> mol*MW = 3.7305/2*63.546 = 118.52 g

c)

how long for I = 10 amp to die...

200 grams...

so

10 hours --> 121.9500 grams consumed

x time --> 200 g

200/121.9500 *10 = 16.400 hours