Originally Posted by joe3335
I don't know what I did wrong. I've tried working out these problems.
Balance the following in acidic solution. (Omit statesofmatter from your answer. Use the lowest possible whole number coefficients.)
(a)
Ti + Cr2O7^2− → Ti^3+ + Cr^3+
Oxidation reaction:
Ti → Ti3+ + 3e
Correct.
Reduction reaction:
Cr2O7^2 → Cr^3+
Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactantside of the equation. Check all the components on the productside of the equation.
Net reaction:
2Ti + Cr2O7^2→ 2Ti^3+ + 2Cr^3+
Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactantside of the equation. Check all the components on the productside of the equation.
Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactantside of the equation. Check all the components on the productside of the equation.

Your reduction half reaction is not balanced
Follow these steps to balance it
1. Balance Cr atoms
2.Balance O by adding water
3.Balance H by adding H+ (acidic medium)
4.Balance charges by adding electrons
You will get this equation
14 H+ +6e^ Cr2O7^2 → 2Cr^3+ 7H2O
Now to add oxidation half reaction with reduction half reactions you need to balance electrons gained and lost
Multiply oxidation half reaction with 2 and add in reduction half reaction
2Ti + Cr2O7^2− + 14 H+ → 2Ti^3+ +2 Cr^3+ + 7H2O
Use same method for other reactions also