# Balancing chem equations under basic conditions?

When the following equation is balanced properly under basic conditions, what are the coefficients of the species shown?

HXeO4- + S2O32 = Xe + SO32-

Water appears in the balanced equation as a --- (reactant, product, neither) with a coefficient of ---. (Enter 0 for neither.)

Which element is oxidized?

Relevance
• Redox reaction .....

Roger has given the balanced net ionic equation .... here's how its done....

HXeO4^- + S2O3^2- --> Xe + SO3^2-

2(HXeO4^- + 3H2O + 6e- --> Xe + 7OH-) ............... reduction half-reaction

3(S2O3^2- + 6OH- --> 2SO3^2- + 3H2O + 4e-) ...... oxidation half-reaction

------------------ -------------------- ------------------- ------------------

2HXeO4^- + 3S2O3^2- + 6H2O + 18OH- --> 2Xe + 6SO3^2- + 14OH- + 9H2O

Simplify

2HXeO4^- + 3S2O3^2- + 4OH- --> 2Xe + 6SO3^2- + 3H2O

• 2 HXeO4{-} + 3 S2O3{2-} + 4 OH{-} → 2 Xe + 6 SO3{2-} + 3 H2O

Water appears in the balanced equation as a product with a coefficient of 3.

S is oxidized.