Balance the following redox reaction occurring in an acidic solution. the coefficient of cr2o72−(aq) is given. enter the coefficients (integers) into the cells before each substance below the equation. 1 cr2o72−(aq) + ti3+(aq) + h+(aq) \rightarrow→ cr3+(aq) + tio2+(aq) + h2o(l)

Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)

Explanation:

Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) +  H⁺ (aq)  → Cr³⁺ (aq) + TiO²⁺ (aq) +  H₂O(l)

This is the reaction, without stoichiometry.

We have to notice the oxidation number of each element.

In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION

Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased.  .- OXIDATION

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.

H₂O  +  Ti³⁺  →  TiO²⁺ + 1e⁻ + 2H⁺

Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.

Now we have to ballance the electrons, so we can cancel them.

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1

(H₂O  +  Ti³⁺  →  TiO²⁺ + 1e⁻ + 2H⁺) .6

Wre multiply, second half reaction .6

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

6H₂O  +  6Ti³⁺  →  6TiO²⁺ + 6e⁻ + 12H⁺

Now we can sum, the half reactions:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻  + 6H₂O  +  6Ti³⁺  → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O

Electrons are cancelled and we can also operate with water and protons

7H₂O - 6H₂O = H₂O

14 H⁺ - 12H⁺ = 2H⁺

The final ballanced equation is:

2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺  → 6TiO²⁺  + 2Cr³⁺ + H₂O

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.