Balance the following redox reaction that takes place in acidic solution: H5IO6(aq) + Cr(s) +   →   IO3- (aq) + Cr3+ (aq) +

To balance the redox reaction, we first need to identify the oxidation and reduction half-reactions.

1. Identify the oxidation and reduction half-reactions:

   • H5IO6(aq) → IO3- (aq) is the reduction half-reaction (Iodine is reduced from +7 to +5)
   • Cr(s) → Cr3+ (aq) is the oxidation half-reaction (Chromium is oxidized from 0 to +3)

2. Balance the atoms and charges in each half-reaction:

   • For the reduction half-reaction:
     • Balance the I atoms: H5IO6(aq) → 2IO3- (aq)
     • Balance the O atoms by adding H2O: H5IO6(aq) → 2IO3- (aq) + 3H2O(l)
     • Balance the H atoms by adding H+: H5IO6(aq) + 10H+ → 2IO3- (aq) + 3H2O(l)
     • Balance the charges by adding electrons: H5IO6(aq) + 10H+ + 2e- → 2IO3- (aq) + 3H2O(l)
   • For the oxidation half-reaction:
     • Balance the Cr atoms: Cr(s) → Cr3+ (aq)
     • Balance the charges by adding electrons: Cr(s) → Cr3+ (aq) + 3e-

3. Equalize the number of electrons in the two half-reactions:

   • Multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2:
     • 3[H5IO6(aq) + 10H+ + 2e- → 2IO3- (aq) + 3H2O(l)]
     • 2[Cr(s) → Cr3+ (aq) + 3e-]

4. Add the half-reactions together and simplify:

   • 3H5IO6(aq) + 30H+ + 6e- + 2Cr(s) → 6IO3- (aq) + 9H2O(l) + 2Cr3+ (aq) + 6e-
   • Cancel out the electrons and simplify the equation to get the final balanced redox reaction:
   • 3H5IO6(aq) + 30H+ + 2Cr(s) → 6IO3- (aq) + 9H2O(l) + 2Cr3+ (aq)

So, the balanced redox reaction in acidic solution is: 3H5IO6(aq) + 30H+ + 2Cr(s) → 6IO3- (aq) + 9H2O(l) + 2Cr3+ (aq)