Question

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Explanation:

Step1: Choose a redox reaction

Let's take the reaction between hydrogen and oxygen to form water: $\ce{2H_{2} + O_{2} \xlongequal{点燃} 2H_{2}O}$. In this reaction, hydrogen is oxidized (oxidation state of H changes from 0 in $\ce{H_{2}}$ to +1 in $\ce{H_{2}O}$) and oxygen is reduced (oxidation state of O changes from 0 in $\ce{O_{2}}$ to -2 in $\ce{H_{2}O}$).

Step2: Determine the oxidation states

• For $\ce{H_{2}}$: The oxidation state of H is 0 (elemental form).
• For $\ce{O_{2}}$: The oxidation state of O is 0 (elemental form).
• For $\ce{H_{2}O}$: The oxidation state of H is +1, and for O is -2 (since the sum of oxidation states in a neutral compound is 0: $2\times(+1) + (-2) = 0$).

Step3: Identify oxidation and reduction

• Oxidation: $\ce{H_{2} -> H_{2}O}$. Each H atom loses an electron (oxidation is loss of electrons). The half - reaction is $\ce{H_{2} -> 2H^{+} + 2e^{-}}$ (in acidic or neutral medium, for simplicity here we can also think in terms of the overall reaction).
• Reduction: $\ce{O_{2} -> H_{2}O}$. Each O atom gains 2 electrons. The half - reaction is $\ce{O_{2} + 4H^{+}+4e^{-} -> 2H_{2}O}$ (in acidic medium, which is relevant for the formation of water from $\ce{H_{2}}$ and $\ce{O_{2}}$ in the presence of an electrolyte or in the context of redox chemistry).
• To balance the electrons, we multiply the oxidation half - reaction by 2: $\ce{2H_{2} -> 4H^{+} + 4e^{-}}$. Now, when we add the two half - reactions (oxidation and reduction), the electrons cancel out:
• Oxidation: $\ce{2H_{2} -> 4H^{+} + 4e^{-}}$
• Reduction: $\ce{O_{2} + 4H^{+}+4e^{-} -> 2H_{2}O}$
• Overall reaction: $\ce{2H_{2} + O_{2} -> 2H_{2}O}$ (which is the same as the reaction we started with, and this shows the redox process).

Answer:

A molecular equation for a redox reaction is $\boldsymbol{\ce{2H_{2} + O_{2} \xlongequal{点燃} 2H_{2}O}}$ (where hydrogen is oxidized and oxygen is reduced).