Question

Balance the following redox (oxidation-reduction) reaction under basic conditions. (Include states-of-matter under the given conditions in your answer. Use the lowest possible whole number coefficients.)
CrO42−(aq) + S2O32−(aq) → Cr3+(aq) + SO42−(aq)

Answer 1

Here us a site that tells you the details.
http://www.chemteam.info/Redox/Redox.html

I don't do them that way. Here is how I do the CrO4^-. You can add the states.
CrO4^2- ==> Cr^3+

Cr is +6 on left and +3 on right. Add 3e to left.
CrO4^2- +3e ==> Cr^3+

Count the charge on the left (-5) and right (+3); add OH^- to the right to balance the charge.
CrO4^2- + 3e ==> Cr^3+ + 8OH^-

Now add H2O to balance.
CrO4^2- + 4H2O + 3e ==> Cr^3+ + 8OH^-