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The redox reaction is:

$\ce{SO2 + Na2CrO4 + H2SO4 -> Na2SO4 + Cr2(SO4)3 + H2O}$

This is what I tried(tho I got stuck at one point which is indeed the exact reason why I've asked this question):

Oxidation half: $\ce{S^4+ -> S^6+ + 2e-}$

Reduction half: $\ce{2Cr^6+ + 6e- -> (Cr^3+)2}$

Net reaction: $\ce{3S^4+ + 2Cr^6+ -> 3S^6+ + (Cr^3+)2}$

Now how do I determine the distribution of $\ce{S^{6+}}$ ions among $\ce{Na_2SO_4}$ and $\ce{Cr_2(SO_4)_3}$??

A similar question to this is:

$\ce{C2H5OH + I2 + OH- -> CHI3 + HCO2- + I- + H2O}$

In the latter one, the distribution of $\ce{C^{2+}}$ among the products is also ambiguous.

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  • $\begingroup$ I don't like the way you wrote the half cells. The overall notion is that when you add the half cells that you get the overall reaction. The point being you can't look up the half cells in a table the way you wrote them. $\endgroup$ – MaxW Jul 29 '18 at 5:54
  • $\begingroup$ @MaxW I've always been taught to write them this way(even the books that I use for practice do it this way)...perhaps you can show me the correct way?? $\endgroup$ – Carrick Jul 29 '18 at 6:08
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    $\begingroup$ I'm used to using a standard electrode potential table as linked and the reduction reaction would be $$\ce{SO4^{2−} + 4 H+ + 2 e− <=> SO2(aq) + 2 H2O }\tag{EMF +0.17}$$ to get oxidation reaction you of course flip equation and sign of EMF. $\endgroup$ – MaxW Jul 29 '18 at 6:16
  • $\begingroup$ @MaxW oh okay...thank you..it is surely the more correct way to represent a half cell rxn..but in my scenario I'm just supposed to solve questions as fast as I can..most of the times its not even necessary to write those half cells while balancing a redox...so we skip most of the details...call it the flaw of education system or something else lol...but that's just how it is $\endgroup$ – Carrick Jul 29 '18 at 6:31
  • $\begingroup$ With your way how do you tell the difference between $$\ce{Cr2O7^{2−} + 14H^+ + 6e− <=> 2Cr^{3+} + 7H2O}$$ and $$\ce{CrO4^{2-} + 4H2O + 3e− <=> Cr(OH)3 + 5 OH^−}$$ $\endgroup$ – MaxW Jul 29 '18 at 6:52
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Well, at first it seems a very complicated equation, but if we remove the soluble ions it gets easier to balance and than we can add then latter. So, this $$\ce{SO2 + Na2CrO4 + H2SO4 -> Na2SO4 + Cr2(SO4)3 + H2O}$$ becomes this: $$\ce{SO2 + CrO4^{-2} -> SO4^{-2} + Cr^{+3}}$$ The sodium ions where there to balance the charge, the sulfuric acid to identify the mean and the water will appear in solution.

Doing this we get that:

The reduction half: $\ce{CrO4^{-2} + 8H+ + 3e- -> Cr^{+3} + 4H2O}$

The oxidation half: $\ce{SO2 + 2H2O -> SO_4^{-2} + 4H+ + 2e-}$

Now multiply each half-cell by an integer necessary to balance the electrons and add the two half cells together. $$\ce{3SO2 + 2CrO4^{-2} + 4H+ -> 3SO4^{-2} + 2Cr^{+3} + 2H2O}$$ Now, we need to add the spectator ions. So first we're going to add sulfate ions. In the process, we're going to join the already existing sulfate and chromium ions. $$\ce{3SO2 + 2CrO4^{-2} + 2H2SO4 -> 2SO4^{-2} + Cr2(SO4)3 + 2H2O}$$ Now, is just add the sodium ions, and we're done: $$\ce{3SO2 + 2Na2CrO4 + 2H2SO4 -> 2Na2SO4 + Cr2(SO4)3 + 2H2O}$$

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  • $\begingroup$ I did not show the betwen steps, because I think you're capable of doing it your self with the simplified equation, but, if you need, I can edit and add the missing steps. A tip that I give to you is balance the charges with electrons after every thing, so you dont need to find any NOX. After this is just a matter of summation of equations. $\endgroup$ – liuzp Jul 29 '18 at 7:13
  • $\begingroup$ damn...that was simple as hell...thank you especially for that second last step...my book had a statement that really confused me: "three $S^{6+}$ ions are distributed in 2 molecules of $Na_2SO_4$ and 1 molecule of $\ce{Cr2(SO4)3}$." $\endgroup$ – Carrick Jul 29 '18 at 7:21
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    $\begingroup$ In the oxidation half cell there should be two waters. // I'd also change "by this you can solve..." to "Now multiple each half-cell by an integer necessary to balance the electrons and add the two half cells together." // Where you have "Now, we need to neutralize the charges." that isn't right. The charges should already be balanced since the half cell charges are balanced. You need to add the "spectator" anions and cations. $\endgroup$ – MaxW Jul 29 '18 at 14:35
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    $\begingroup$ MaxW I'm going to fix it. Thanks. As I'm not fluent speeker, a few things can be misspeled. And in the neutralization step, I was talking about the charges present in both sides, and not any unbalanced one, but I'm going to change for something more comprehensive. $\endgroup$ – liuzp Jul 29 '18 at 15:48
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    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. The mhchem construct \ce{...} makes writing chemical equations much easier, please have a look here and here. $\endgroup$ – Martin - マーチン Aug 2 '18 at 18:43

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