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Use the image again and determine which elements have been oxidized, reduced, or unchanged. (4 points) Oxidized Reduced Unchanged The (blue) oxygen atom The (green) hydrogen atom The (red) oxygen atom The (red) carbon atom c(+4)o(-2)2+h(+1)2o(-2)-- c(?)h(+1)2o(-2)+0(0)2

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To determine which elements have been oxidized, reduced, or unchanged in the reaction, we first need to identify the oxidation states of the elements involved in the given equations.

For the reactants:

  1. C (Carbon) in the compound \( \text{C}(\text{+4}) \).
  2. H (Hydrogen) in \( \text{H}_2\text{O} \): each \( \text{H} \) has an oxidation state of \( \text{+1} \).
  3. O (Oxygen) in \( \text{H}_2\text{O} \): each \( \text{O} \) has an oxidation state of \( \text{-2} \).

For the products:

  1. C (Carbon): If the carbon is unchanged, its oxidation state would still be \( \text{+4} \).
  2. H (Hydrogen): It remains \( \text{+1} \) in \( \text{H}_2\text{O} \).
  3. O (Oxygen): In \( \text{O}_2 \), the oxidation state is \( 0 \).

Analyzing the changes:

  • The carbon (C) remains at \( +4 \) during the reaction, so it is unchanged.
  • The hydrogens (H) remain at \( +1 \), so they are also unchanged.
  • The red oxygen atom in \( H_2O \) (with \(-2\)) is converted to \( O_2 \) (with \( 0 \)), indicating oxidation (loss of electrons).
  • The blue oxygen atom in the \( O_2 \) could be said to not get reduced because it is forming a molecule from elemental oxygen, hence no change in oxidation state occurs for it but rather it combines.

To summarize:

  • Oxidized: The red oxygen atom (going from \(-2\) in water to \(0\) in \(O_2\))
  • Reduced: None
  • Unchanged: The carbon atom and hydrogen atom (both remain in the same oxidation states: carbon at \(+4\) and hydrogen at \(+1\)).

In conclusion:

  • Oxidized: Red Oxygen Atom
  • Reduced: None
  • Unchanged: Carbon Atom, Green Hydrogen Atom