In acidic solution: Complete and balance the above half-reaction. In each case indicate whether the half-reaction is an oxidation or a reduction. Thank you!
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,
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
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