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I have got this redox reaction:

$$\ce{(NH_4)_2Cr_2O_7 + H_2SO_4 + Na_2SO_3 + H_2O -> Cr_2(SO_4)_3 + (NH_4)_2SO_4 + NaOH}$$

I need to balance it. I have no problem with finding what elements change their oxidation number.

$$\ce{Cr^6+ + 3e^- -> Cr^3+}$$

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

The problem here for me is that there is sulfur in two different compounds on both sides of reaction so I do not know where to fill it and if I try it never goes well.

I am from Czech Republic so we maybe use different method that does not apply for those more complicated equations, I am of course willing to learn yours if it will solve the case. Thanks for any feedback!

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  • $\begingroup$ Having $\ce{H2SO4}$ on the left simultaneously with $\ce{NaOH}$ on the right makes no sense at all. $\endgroup$ – Ivan Neretin Mar 4 '16 at 12:49
  • $\begingroup$ @IvanNeretin this is an equation from my chemistry teacher, it could be that she is wrong and this equation can not happen, I do not know $\endgroup$ – Vojta Klimes Mar 4 '16 at 13:57
  • $\begingroup$ Why, it can, only the products would be somewhat different (note $\ce{Na2SO4}$ in the answer by Yomen Atassi). $\endgroup$ – Ivan Neretin Mar 4 '16 at 14:02
  • $\begingroup$ Oh, I actually overlooked it anyway. $\endgroup$ – Vojta Klimes Mar 4 '16 at 14:09
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Let's begin by the two half redox equations: $$\ce{14H+ + Cr2O7^{2-} + 6e- <=> 2Cr^{3+} + 7 H2O}$$ $$\ce{SO3^{2-} + H2O <=> SO4^{2-} +2e- +2H+}$$ By multiplying the second equation by 3 and adding the two equations, we find: $$\ce{8H+ + Cr2O7^{2-} + 3SO3^{2-} + <=> 2Cr^{3+} + 3SO4^{2-} + 4 H2O}$$ We add to the two sides of the last equation: $\ce{4SO4^{2-}, 2NH4+ , 6Na+}$ and we rearrange the equation: $$\ce{4H2SO4 + (NH4)_2Cr2O7 + 3Na_{2}SO3 <=> Cr_{2}(SO4)_3 + 3Na_2SO4 + (NH4)_2SO4 + 4 H2O}$$

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  • $\begingroup$ Ok it is mostly clear now, just I do not know how you know what elements you give to the first and second half redox equations. $\endgroup$ – Vojta Klimes Mar 3 '16 at 19:18
  • $\begingroup$ Would you please precise your question...what do you mean what elements you give to the first and second half redox equations? You precised in your questions the elements that were oxidized or reduced. I only balanced the equations. $\endgroup$ – Yomen Atassi Mar 4 '16 at 6:45
  • $\begingroup$ I just do not know how to know that you just put H+ and Cr2O7 and so on to the first equation, what is the reason for you to put those elements there. $\endgroup$ – Vojta Klimes Mar 4 '16 at 13:56
  • $\begingroup$ $\ce{H+}$ is because we work in acidic medium, and $e^-$ is because we need to reduce that bichromate. $\endgroup$ – Ivan Neretin Mar 4 '16 at 13:59
  • $\begingroup$ I agree with @IvanNeretin, we usually balance the charge of half redox equation with either $\ce{H+}$ or $\ce{OH-}$ depending on the pH of the medium and we balance the matter with $\ce{H2O}$ $\endgroup$ – Yomen Atassi Mar 4 '16 at 14:11

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