Balancing a Redox Equation with Half-Reactions

[RaamGeneral]

It's not clear to me how I can balance the following equation, most of all because of K which I don't know how to deal with:

\mathrm{SO_2 + K_2Cr_2O_7 \to Cr_2(SO_4)_3 + SO_3} [this is the text of the exercise]

It's acidic solution for sulfuric acid, so the equation can be better written so:

\mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4 \to Cr_2(SO_4)_3 + SO_3 + H_2O + K} (I assume K has to be added, preserving the charge in both members)

To use the half-reaction method I write the ionic form:

\mathrm{SO_2+K^+ + Cr_2O_7^{2-}+H^+ \to Cr^{3+}+SO_3+H_2O + K}

Writing and balancing three half-reactions (can it be?) I get:

\mathrm{4SO_2 + K_2Cr_2O_7+3H_2SO_4 \to 4SO_3 + Cr_2(SO_4)_3+2K+3H_2O}Is it correct?

Thank you.

 

[Borek]

Hint: K⁺ is just a spectator, and SO₃ in water means just SO₄^2-.

 

[RaamGeneral]

Ok, your hint was foundamental to me.
I have already thought about K+ being spectator but I "forgot" that
\mathrm{ SO_3 + H_2O\to H_2SO_4\to 2H^+ + SO_4^{2-} } (right?) so I didn't know what to do without K in the second member.\mathrm{SO_2 + K_2Cr_2O_7 + H_2SO_4\to Cr_2(SO_4)_3 + SO_3 + H_2O + 2K^+ }
\mathrm{SO_2 + K_2Cr_2O_7 \to Cr_2(SO_4)_3 + 2K^+ }
ionic form: \mathrm{SO_2 + Cr_2O_7^{2-}\to Cr^{3+} + SO_4^{2-} }

Balancing the two half-reactions:

\mathrm{3SO_2 + Cr_2O_7^{2-} + 2H^+\to 3SO_4^{2-} + 2Cr^{3+} + H_2O }

And finally:

\mathrm{3SO_2 + K_2Cr_2O_7 + H_2SO_4\to K_2SO_4 +Cr_2(SO_4)_3 + H_2O }I'm not so sure though. Is it correct?

Thank you for your reply.

 

[Borek]

Looks OK to me.