Part A:
Balance each of the following redox reactions occurring in acidic aqueous solution.
MnO4−(aq)+Al(s)→Mn2+(aq)+Al3+(aq)
Part B:
What mass of lead sulfate is formed in a lead-acid storage battery when 1.40 g of Pb undergoes oxidation?
Part C:
A meteor has a Pb−206:U−238 mass ratio of 0.840:1.00.
What is the age of the meteor? (Assume that the meteor did not contain any Pb−206 at the the time of its formation.)
Part A:
Balancing in acidic medium
The reduction equation will be:
MnO4- --> Mn+2
We will balance oxygen by adding water molecules and then balance hydrogen by adding H+
MnO4- + 8H+ --> Mn+2 + 4H2O
Now balance charge
MnO4- + 8H+ + 5e --> Mn+2 + 4H2O ...................1
oxidation reaction is
Al --> Al+3 , we need to balance charge only
Al --> Al+3 + 3e....................2
Now we will add the equation by multiplying with suitable coeffeceients so that charge will be balanced
3MnO4- + 24H+ + 5Al --> 3Mn+2 + 12H2O + 5Al+3
Part B
The entire redox reaction in lead acid storage battery is
Pb(s) + PbO2(s) + 2H2SO4(aq) --> 2PbSO4(s) + 2H2O(l)
Molecular weight of PbSO4= 303.3 g / mole
Atmoc weight of Pb =207.2 g /mole
As per the balance equation
1 mole of Pb will give 2 moles of PbSO4
so 207.2 grams will give 2X303.3 grams of PbSO4
So 1 gram will give 606.6 / 207.2 grams of PbSO4
so 1.4 gram will give 606.6 X 1.4 / 207.2 grams = 4.102 grams
Part C
As there is no Pb-206 at the start so it is formed only by radioactive decay of uranium
The half-life of uranium decay is 4.47 billion years.,
we know tha t1/2 = 0.693 / K
So K = disintegration constant = 0.155 billion years -1
Now for first order equation
N(t) = N(0) e-Kt
N(t)=Number of uranium atoms at time t
N(0)=Number of uranium atoms initally =N(t)+N(Pb)=sum of uranium and Pb atoms after time t.
N(t)=(N(t)+N(Pb)) e^-kt
or, N(t)/N(t)+N(Pb)=e^-Kt
1+ N(Pb)/N(t)=e^-Kt
N(Pb)/N(t)=e^-Kt-1..........(1)
We have mass ration of Pb / U = 0.840 / 1.0
We know that mole = mass / molecular weight so
Mole ratio = 0.840/1.0 X 238/206 ( atomic weight of Uranium = 238 and Pb = 206)
=0.97/1.00 * NA/NA=N(Pb)/N(t)=0.97/1.00 (NA is avogadro's no)
from eqn (1), 0.97/1.00=e^-kt-1
1.97=e-0.155t
0.678=0.155t
t=4.374 billion years.
This will be the age of meteor
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