A small dry-cell battery of zinc, ammonium chloride, and MnO2 weighing 85 g will
ID: 905426 • Letter: A
Question
A small dry-cell battery of zinc, ammonium chloride, and MnO2 weighing 85 g will operate continuously through a 4-ohm resistance for 450 min before its voltage falls below 0.75V. The initial voltage is 1.60, and the effective voltage over the whole life of the battery is taken to be 1.00V. How much work is obtained from the battery, in Joules? How many moles of Zn are converted to ZnCl2 in this process, and what is the weight of this amount of ZnCl2? How much MnO2 is required to carry out this oxidation? What fraction of the total 85-g weight of the battery consists of chemical reagents? About how many kilometers above the earth’s surface could this 85-g battery be raised with this amount of work? (Hints: Remember that the current pushed by a battery through a resistance R is given by I=V/R. The power it generates is P=IV=V2/R, and the total work it performs over a period of time is given by w =V 0 I(t) dt VQ = VI = V2/R. Q is the total charge that the battery pushes through a roughly-constant voltage. So the essence of this problem is to calculate the values of w and Q, the total electrical charge that can be generated by the battery, and then to assume that all that charge is generated by the conversion of a certain amount of Zn and Cl- to ZnCl2).
Explanation / Answer
A small dry-cell battery of zinc, ammonium chloride, and MnO2 weighing 85 g will operate continuously
through a 4-ohm resistance for 450 min before its voltage falls below 0.75V.
The initial voltage is 1.60, and the effective voltage over the whole life of the battery is taken to be
1.00V.
R = 2 ohm
Vi = 1.6 V
Vf = 0.75 V
Veff = 1.0 V
time, = 450 min = 450*60 = 27000 s
(i) How much work is obtained from the battery, in Joules?
work done, w = *Veff^2/R = 27000*1^2/2 = 13500 J = 13.5 kJ
(ii) How many moles of Zn are converted to ZnCl2 in this process, and what is the weight of this amount
of ZnCl2?
Amounts of charge transferrred, Q = I* = *Veff/R = 27000*1.0/2 = 13500 C
Charge of 1 mole of elctrons = 1F = 96500 C
No of moles of electrons transformed, n = 13500/96500 = 0.14 mol
Anode: Zn(s) Zn2+(aq) + 2 e-
2 moles of electrons convert 1 mol of Zn
0.14 moles of electrons will convert 0.14*1/2=0.07 moles of Zn
So 0.07 moles of Zn will be converted and 0.07 moles of ZnCl2 will be produced.
Molar mass of ZnCl2 = 136.315 g/mol
Mass of 0.07 moles of ZnCl2 = 0.07*136.315 = 9.542 g
(iii) How much MnO2 is required to carry out this oxidation?
Cathode: 2MnO2 + 2NH4+ + 2e- Mn2O3 + 2NH3 + H2O
2 moles of electrons require 2 mols of MnO2
0.14 moles of electrons will convert 0.14*2/2=0.14 moles of MnO2
So 0.14 moles of MnO2 will be required.
Molar mass of MnO2 = 86.94 g/mol
Mass of 0.14 moles of MnO2 = 0.14*86.94 = 12.17 g
(iv) What fraction of the total 85-g weight of the battery consists of chemical reagents?
Amount of NH4Cl required = 0.14 mol
Molar mass of NH4Cl = 53.5 g/mol
Mass of 0.14 moles of NH4Cl = 0.14*53.5 = 7.5 g
Molar mass of Zn = 65.4 g/mol
Mass of 0.07 moles of Zn required = 0.07*65.4 = 4.6 g
Mass of 0.14 moles of MnO2 = 0.14*86.94 = 12.2 g
Total mass of reagents = 7.5+4.6+12.2 = 24.3 g
Fraction of total weight = 24.3/85 = 0.28
(v) About how many kilometers above the earth’s surface could this 85-g battery be raised
with this amount of work?
mass, W = 85 g = 0.085 kg
w = 13500 J = mass*9.8*h = 0.085*9.8*h
h = 13500/(0.085*9.8) = 16206 m = 16.21 km