What is the half reaction for TiO2(s) –> Ti2+(aq)?

SOMEONE ASKED 👇

What is the half reaction for TiO2(s) –> Ti2+(aq)?

HERE THE ANSWERS 👇

• Ti O2(s) —-> Ti2+(aq)

  +4 -4 ———>+1 , since Ti went from +4 to +1 (the electrons became more negative), this is a reduction.

  Step1: Balance everything except for hydrogen and oxygen. It is already balanced.

  Ti (s) ——-> Ti2+(aq)

  Step 2: Balance the Oxygen in the equation by adding a H2O molecular for each O atom, we have 2 in this equation.

  TiO2(s) —–> Ti2+(aq) + 2H2O(l)

  Step 3: Balance the Hydrogen by adding hydrogen ion, H+ to the appropriate side.

  TiO2(s) + 4H(g) ——> Ti2+(aq) + 2H2O(l)

  Step 4: Balance charge by adding electrons, e-, to the appropriate side. Since, Ti when from +4 to a +1, there is a difference of 3, so we add 3e- on the reaction side of the equation.

  TiO2(s) + 4H(g) + 3e- ——> Ti2+(aq) + 2H2O(l), this is your answer.

  Hope this helped,

  Bill

  Source(s): q

• You’re right, There’s +4 from hydrogen on the left side and 2+ on the right side from Ti. so you only need 2 electrons

• but how tio2 forms oxigen vacancies ? and how used for resistive switching?? if any method to create oxigen vacancies without anneling or temp application? ty n advance