Chapter 17: Problem 51

Write the balanced half-reaction for the conversion of \(\mathrm{H}_{2} \mathrm{O}(l)\) to \(\mathrm{O}_{2}(a q)\) in acidic solution.

Short Answer

Expert verified

Balanced half-reaction for the conversion of liquid water to aqueous oxygen in acidic solution is: \[2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(a q) + 4\mathrm{H}^+ + 4\mathrm{e}^-\]

Step by step solution

Write The Unbalanced Equation

Initially, write down the unbalanced reaction as given: \[\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(a q)\]

Balance Oxygen Atoms

Balance the oxygen atoms by adding water molecules. There are 2 oxygen atoms on the right side, so we need 2 \(\mathrm{H}_{2}\mathrm{O}(l)\) on the left: \[2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(a q)\]

Balance Hydrogen Atoms

Balance the hydrogen atoms by adding 4 protons (H\(^+\)) on the right side. There are 4 hydrogen atoms on the left-side, so balance that by adding 4 protons on the right side: \[2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(a q) + 4\mathrm{H}^+\]

Balance Charge

Balance the charges by adding electrons. As we have 4 positive charges owing to protons on the right, add 4 electrons on the right side to neutralize it: \[2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{O}_{2}(a q) + 4\mathrm{H}^+ + 4\mathrm{e}^-\]

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Write the balanced half-reaction for the conversion of \(\mathrm{H}_{2} \mathrm{O}(l)\) to \(\mathrm{O}_{2}(a q)\) in acidic solution.

Short Answer

Step by step solution

Write The Unbalanced Equation

Balance Oxygen Atoms

Balance Hydrogen Atoms

Balance Charge

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