Question

Which of the following is not a redox reaction? (a) \(\mathrm{CuO}+\mathrm{H}_{2} \rightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Na}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH}+\frac{1}{2} \mathrm{H}_{2}\) (c) \(\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}+\mathrm{CO}_{2}\) (d) \(2 \mathrm{~K}+\mathrm{F}_{2} \rightarrow 2 \mathrm{KF}\)

Short answer

Reaction (c) is not a redox reaction.

Step-by-step solution

Identify Oxidation and Reduction

Check each reaction to see if there is a change in oxidation states of the elements involved, which would indicate electron transfer. In redox reactions, one species is oxidized (loses electrons) and another is reduced (gains electrons).

Analyze Reaction (a)

In reaction (a), \( \mathrm{CuO} + \mathrm{H}_2 \rightarrow \mathrm{Cu} + \mathrm{H}_2\mathrm{O} \), copper (Cu) is reduced from a +2 to a 0 oxidation state and hydrogen (H) is oxidized from 0 to +1. Both reduction and oxidation occur.

Analyze Reaction (b)

In reaction (b), \( \mathrm{Na} + \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{NaOH} + \frac{1}{2} \mathrm{H}_2 \), sodium (Na) is oxidized from 0 to +1 and hydrogen (H) in water is reduced from +1 to 0. Both reduction and oxidation occur.

Analyze Reaction (c)

In reaction (c), \( \mathrm{CaCO}_3 \rightarrow \mathrm{CaO} + \mathrm{CO}_2 \), calcium (Ca) remains with an oxidation state of +2 in both \( \mathrm{CaCO}_3 \) and \( \mathrm{CaO} \) and carbon (C) does not change its oxidation state as it is in the form of a carbonate and evolves as carbon dioxide. There is no transfer of electrons.

Analyze Reaction (d)

In reaction (d), \( 2 \mathrm{K} + \mathrm{F}_2 \rightarrow 2 \mathrm{KF} \), potassium (K) is oxidized from 0 to +1 and fluorine (F) is reduced from 0 to -1. Both reduction and oxidation occur.

Determine the Non-Redox Reaction

Since reaction (c) does not involve any change in oxidation states, it is not a redox reaction.

Key concepts

Oxidation and Reduction

Understanding the dance between oxidation and reduction is crucial for grasping redox reactions in chemistry. Oxidation is like a partner in this dance who gives away electrons, and as a result, its oxidation state increases. This is because electrons are negatively charged, so losing electrons makes an atom more positive. Reduction, on the other hand, is the process where another partner gains those electrons, decreasing its oxidation state due to the increased negative charge.

Oxidation and reduction always occur together; when one species gives up electrons, another must accept them. This is why we often speak of 'redox pairs' and use the mnemonic 'OIL RIG' which stands for 'Oxidation Is Loss, Reduction Is Gain.' These processes are fundamental to a plethora of chemical reactions, including those that power batteries, and even the metabolic processes within our bodies.

In the exercise, we can see that reactions (a), (b), and (d) involve changes in oxidation states and hence are true redox reactions. To decipher this, one must carefully observe the initial and final oxidation states of the elements in a chemical reaction. If one species is oxidized and another is reduced, it's a clear indicator that electrons have been transferred, confirming the reaction is indeed a redox reaction.

Oxidation States

Oxidation states, also commonly referred to as oxidation numbers, are a way to keep 'score' of the electrons when atoms in molecules or ions share electrons. They tell us about the degree of oxidation of an atom in a chemical compound, which can be zero for elements, positive or negative based on the rules of assigning oxidation states. These rules rely on the assumption that the more electronegative element takes the shared electrons, which is akin to assuming 'what if the bond was ionic?'

To assign oxidation states, we use known values for elemental states, typically zero, and known values for common ions, like +1 for sodium. But, when looking at compounds, some basic rules come into play: for example, oxygen often has an oxidation state of -2 and hydrogen +1, except when it forms compounds with metals where it is -1. The sum of the oxidation states in a neutral molecule must be zero, or in ions, equal to the charge of the ion.

The analysis of oxidation states in the provided exercise enables us to identify that reaction (c) lacks the electron transfer characteristic to a redox reaction because the oxidation states of calcium and carbon remain unchanged before and after the reaction.

Chemical Reactions Analysis

Chemical reactions analysis is like being a detective; you examine the evidence, the reactants, and products, to understand the storyline of how atoms rearrange themselves during chemical reactions. One aspect of this analysis is observing the conservation of mass and charge, ensuring that atoms aren't magically appearing or disappearing as we write balanced chemical equations. Another aspect is identifying reaction types: synthesis, decomposition, single replacement, double replacement, or combustion. And crucially for this topic, determining whether a reaction involves oxidation and reduction.

By investigating the oxidation states of the elements before and after the reaction, as done in the step-by-step solution for the exercise, we detect whether a redox process has occurred. Reaction (c) stands out because it is a decomposition reaction, breaking down calcium carbonate to calcium oxide and carbon dioxide without any change in oxidation states, thus signaling that it's not a redox reaction. This analysis is vital for understanding the changes happening at the molecular level, which can be useful in various applications, from industrial synthesis to environmental chemistry.