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?

2 Answers

Relevance
  • 1 year ago

    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

  • 1 year ago

    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.

Still have questions? Get your answers by asking now.