-2
$\begingroup$

I've started balancing redox reactions, and one particular reaction has tripped me up.

$$\ce{PbS + O3 -> PbSO4 + O2}$$

$\Delta ON$ for $S =8$

$\Delta ON$ for $O =-2$

Here, $O_2$ acts as a spectator ion as its oxidation number remains $0$, and $PbSO_4$ is the RHS in both half reactions as its sulphur and oxygen get oxidised and reduced, respectively.

Writing the half reactions:

$$ \bigg( \,\, \ce{PbS -> PbSO4} \,\,\,\,\,\,\,\,\,\,\,\, \Delta ON = 8 \,\, \bigg) \times 3$$

$$ \bigg( \,\, \ce{4O3 -> 3PbSO4} \,\,\,\,\,\,\,\,\,\,\,\, \Delta ON = -12 \,\, \bigg) \times 2$$

Adding the two reactions:

$$\ce{3PbS + 8O3 -> 9PbSO4}$$

This is clearly wrong, as in this case, we would have to balance the $O_2$ spectator ion on the LHS, while it is clearly in the RHS in the original reaction, and upon further researching online, I found the actual balanced reaction:

$$\ce{PbS + 4O3 -> PbSO4 + 4O2}$$,

This indicating my half - reactions itself were wrong.

Upon doing some research online, it turns out this reaction utilises the n - factor concept, which I'm not very familiar with. If someone could point out where I'm going wrong in this concept, I'd be very grateful.

$\endgroup$
0

2 Answers 2

1
$\begingroup$

$\ce{O2}$ is not a spectator ion, as it is neither a ion nor a spectator. It is a reaction product.

You can formally separate the reaction to

$$\ce{4 O3 -> 4 O2 + 4 O}$$

with $\Delta \text{ON} = 0$ and

$$\ce{PbS + 4 O -> PbSO4}$$

with

  • $\Delta \text{ON}_\ce{S} = (+6)-(-2))= +8$
  • $\Delta \text{ON}_\ce{O} = 4 \cdot ((-2) - (0)) = 4 \cdot (-2) = -8$

You can effectively imagine the $\ce{O3}$ molecule as $\ce{O2}$ carrying an extra $\ce{O}$ as the oxidation agent.

Similar reaction, used for restoration of old painting where lead white ($\ce{2PbCO3·Pb(OH)2}$) was used is:

$$\ce{PbS + 4 H2O2 -> PbSO4 + 4 H2O}$$

$\endgroup$
1
  • $\begingroup$ Thank you for the answer. Could you please state the $\Delta ON$ for both individual half reactions? $\endgroup$
    – Mr Gubbo
    Feb 6 at 14:55
0
$\begingroup$

This equation cannot be balanced unless one is given the exact stoichiometry and the mechanism is known. It might be reasonable to think that O2 could oxidize PbS:

PbS + 2O2 = PbSO4 and knowing 2O3 = 3O2 get PbS + 2O3 = PbSO4 + O2; Perfectly reasonable and balanced.

Were one told that one mole of PbS required 4 moles of O3 and produced 4 moles of O2 then the equation would simply be PbS + 4O3 = PbSO4 + 4O2. The reaction is not amenable to balancing by the half reaction method.

Should one remember high school chemistry that sulfide ores are roasted in air [O2] producing oxides of the metal and SO2 one might think that there was something special about ozone oxidizing sulfide to sulfate and jump directly to the second equation. There is more to chemistry than just balancing an equation. The equation is telling something!

$\endgroup$
1
  • $\begingroup$ Never thought of it that way. Thank you for your answer. $\endgroup$
    – Mr Gubbo
    Feb 7 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.