Questions: Study this chemical reaction: Ti + O2 -> TiO2 Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction. oxidation: reduction:

Study this chemical reaction: Ti + O2 -> TiO2 Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction. oxidation: reduction:
Transcript text: Study this chemical reaction: \[ \mathrm{Ti}+\mathrm{O}_{2} \rightarrow \mathrm{TiO}_{2} \] Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction. oxidation: reduction:
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Solution

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Solution Steps

Step 1: Identify the Oxidation and Reduction Processes

In the given reaction, titanium (Ti) is reacting with oxygen (\(\mathrm{O}_2\)) to form titanium dioxide (\(\mathrm{TiO}_2\)). To write the balanced half-reactions, we need to identify which element is oxidized and which is reduced.

  • Oxidation involves the loss of electrons.
  • Reduction involves the gain of electrons.
Step 2: Determine Oxidation States
  • In the reactants, titanium (\(\mathrm{Ti}\)) is in its elemental form, so its oxidation state is 0.
  • In \(\mathrm{TiO}_2\), titanium is typically in the +4 oxidation state.
  • Oxygen (\(\mathrm{O}_2\)) in its elemental form has an oxidation state of 0.
  • In \(\mathrm{TiO}_2\), oxygen is typically in the -2 oxidation state.
Step 3: Write the Oxidation Half-Reaction

Since titanium goes from an oxidation state of 0 to +4, it is oxidized. The oxidation half-reaction is:

\[ \mathrm{Ti} \rightarrow \mathrm{Ti}^{4+} + 4e^{-} \]

Step 4: Write the Reduction Half-Reaction

Oxygen goes from an oxidation state of 0 to -2. Since each oxygen atom gains 2 electrons, and there are two oxygen atoms in \(\mathrm{O}_2\), the reduction half-reaction is:

\[ \mathrm{O}_2 + 4e^{-} \rightarrow 2\mathrm{O}^{2-} \]

Final Answer

  • Oxidation half-reaction:

    \[ \boxed{\mathrm{Ti} \rightarrow \mathrm{Ti}^{4+} + 4e^{-}} \]

  • Reduction half-reaction:

    \[ \boxed{\mathrm{O}_2 + 4e^{-} \rightarrow 2\mathrm{O}^{2-}} \]

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