Which one of the following is an oxidation-reduction reaction? naoh + hno3 --> h2o + kno3 naoh + hno3 --> h2o + kno3 so3 + h2o --> h2so4 cacl2 + na2co3 --> caco3 + 2 nacl ch4 + 2 o2 --> co2 + 2 h2o al2(so4)3 + 6 koh --> 2 al(oh)3 + 3 k2so4

CH4 + 2 O2 --> CO2 + 2 H2O

Explanation:

CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...

O₂ → CO₂ + 4e⁻ ==> oxidation

O₂ + 4e⁻  →  H₂O ==> reduction

2O₂ + 4e⁻  →  CO₂ + H₂O + 4e⁻  ==> Net oxidation-reduction

=>  4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.

Then...

Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...

2CH₄ + 2O₂ + 8e⁻  →  2CO₂ + 2H₂O + 8e⁻

Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...

2CH₄ + 2O₂  →  2CO₂ + 2H₂O

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O  

Explanation:

To identify an oxidation-reduction reaction, you must determine the oxidation number of every atom involved in the reaction and see if it changes.

The only reaction where two elements change oxidation number is the oxidation of methane.

Here's the oxidation number of every atom involved:.

We see that some elements change oxidation numbers.

C:  -4 ⟶ +4; increase in oxidation number = oxidation

O:  0 ⟶  -2; decrease in oxidation number = reduction

H: +1 ⟶   +1; no change.

The reaction is an oxidation-reduction reaction, because carbon is oxidized, and oxygen is reduced.

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1 a

2 d

3 a

4 c

explanation: