# How to balance this redox reaction?

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!

• Having $\ce{H2SO4}$ on the left simultaneously with $\ce{NaOH}$ on the right makes no sense at all. – Ivan Neretin Mar 4 '16 at 12:49
• @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 – Vojta Klimes Mar 4 '16 at 13:57
• Why, it can, only the products would be somewhat different (note $\ce{Na2SO4}$ in the answer by Yomen Atassi). – Ivan Neretin Mar 4 '16 at 14:02
• Oh, I actually overlooked it anyway. – Vojta Klimes Mar 4 '16 at 14:09

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: $$\ce{4H2SO4 + (NH4)_2Cr2O7 + 3Na_{2}SO3 <=> Cr_{2}(SO4)_3 + 3Na_2SO4 + (NH4)_2SO4 + 4 H2O}$$
• $\ce{H+}$ is because we work in acidic medium, and $e^-$ is because we need to reduce that bichromate. – Ivan Neretin Mar 4 '16 at 13:59
• 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 $\ce{H2O}$ – Yomen Atassi Mar 4 '16 at 14:11