Question

A rechargeable battery is constructed based on a concentration cell constructed of two Ag/Ag+ half-cells. The volume of each half-cell is 1.8 L and the concentrations of Ag+ in the half-cells are 1.35 M and 1.0×10⁻³ M .

a) For how long can this battery deliver 2.8 A of current before it goes dead?

b) What mass of silver is plated onto the cathode by running at 3.8 A for 5.7 h ?

c) Upon recharging, how long would it take to redissolve 110 g of silver at a charging current of 10.0 amps?

Answer 1

(a)
two half reactions of the cells are:

Cathode
Ag(+) + e(-) ----> Ag

Anode
Ag -----> Ag(+) + e(-)

The voltage becomes 0 after the concentrations become equal. The final concentrations

C = (C1 + C2)/2 = (1.35 + 1*10^-3)/2 = 0.6755 mole/L.

Moles of Ag(+) ions

=(1.35 - 0.6755)*1.8 = 1.2141 moles

Then:
Q = moles x coulamb/moles

= 1.2141*96500 = 117160.65 C
Time = 117160.65 C/(2.8 A) = 46864.24 s
40289 (s)/(60s*60min)=13 hr

(b). Mass of Silver = current x time x MW / coulamb x M

= 3.8A*20520s*(107.87g/molAg)/96500

= 87.16 g

(c). Time to redissolve

= (moles x coulamb/moles ) / current

= (110g/(107.87g/molAg)*96500) / (10)
= 9840.549/(60s*60min)

=2.73hr