S changes from +4 in SO3^-2 to +6 in SO4^-2, a loss of 2 electrons.
Mn changes from +7 in MnO4^- to +2 in Mn^+2, a gain of 5 electrons.
Multiply the Mn half reaction x 2.
Multiply the SO3^-2 half reaction x 5.
Count up oxygen atoms and balance H2O, then add H^+ to the left side to balance H atoms. Check that the charge balances.
Balance the following oxidation-reduction reaction in acid solution:
SO32- + MnO4- --> SO42- + Mn2+ + H2O
2 answers
What is the pH ( 3 significant digits) of a 0.01 M solution of sulfuric acid? Hint is that the first ionization constant is "large", and the second is finite, and the [H+] donated by the second equilibrium depends upon the amount of [H+] already donated by the first ionization. What is the answer please.
1 answer