Problem 17 - Chemistry The Central Science In Si Units

Question

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) \(14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)\) $$ \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q) $$ (b) \(2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) $$ \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q) $$ (c) \(\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Step-by-Step Solution

Answer

For reaction (a), 10 electrons are transferred; for (b), 6 electrons; for (c), 2 electrons.

Step 1: Assign Oxidation Numbers for Reaction (a)

For reaction (a):- In the reactants: - H in \(\mathrm{H}^{+}(aq)\) is \(+1\). - Mn in \(\mathrm{Mn}^{2+}(aq)\) is \(+2\). - Na in \(\mathrm{NaBiO}_3(s)\) is \(+1\), Bi is \(+5\), and O is \(-2\).- In the products: - H in \(\mathrm{H}_{2}O(l)\) is \(+1\), O is \(-2\). - Mn in \(\mathrm{MnO}_4^{-}\) is \(+7\), O is \(-2\). - Bi in \(\mathrm{Bi}^{3+}(aq)\) is \(+3\). - Na in \(\mathrm{Na}^{+}(aq)\) is \(+1\).

Step 2: Determine Electrons Transferred for Reaction (a)

- Mn is oxidized from \(+2\) in \(\mathrm{Mn}^{2+}\) to \(+7\) in \(\mathrm{MnO}_4^{-}\), losing 5 electrons per Mn.- Bi is reduced from \(+5\) in \(\mathrm{NaBiO}_3\) to \(+3\) in \(\mathrm{Bi}^{3+}\), gaining 2 electrons per Bi.- Total electrons transferred is 10 (since 2 Mn lose 10 electrons and 5 Bi gain 10).

Step 3: Assign Oxidation Numbers for Reaction (b)

For reaction (b):- In the reactants: - K in \(\mathrm{KMnO}_4\) is \(+1\), Mn is \(+7\), and O is \(-2\). - Na in \(\mathrm{Na}_2\mathrm{SO}_3\) is \(+1\), S is \(+4\), and O is \(-2\).- In the products: - Mn in \(\mathrm{MnO}_2\) is \(+4\), O is \(-2\). - Na in \(\mathrm{Na}_2\mathrm{SO}_4\) is \(+1\), S is \(+6\), and O is \(-2\). - K in \(\mathrm{KOH}\) is \(+1\), O is \(-2\), H is \(+1\).

Step 4: Determine Electrons Transferred for Reaction (b)

- Mn is reduced from \(+7\) in \(\mathrm{KMnO}_4\) to \(+4\) in \(\mathrm{MnO}_2\), gaining 3 electrons per Mn.- S is oxidized from \(+4\) in \(\mathrm{Na}_2\mathrm{SO}_3\) to \(+6\) in \(\mathrm{Na}_2\mathrm{SO}_4\), losing 2 electrons per S.- Total electrons transferred is 6 (since 2 Mn gain 6 electrons and 3 S lose 6).

Step 5: Assign Oxidation Numbers for Reaction (c)

For reaction (c):- In the reactants: - Cu is \(0\) in \(\mathrm{Cu}(s)\). - Ag in \(\mathrm{AgNO}_3\) is \(+1\), N is \(+5\), and O is \(-2\).- In the products: - Ag is \(0\) in \(\mathrm{Ag}(s)\). - Cu in \(\mathrm{Cu(NO}_3)_2\) is \(+2\).

Step 6: Determine Electrons Transferred for Reaction (c)

- Cu is oxidized from 0 in \(\mathrm{Cu}(s)\) to \(+2\) in \(\mathrm{Cu(NO}_3)_2\), losing 2 electrons.- Each Ag is reduced from \(+1\) in \(\mathrm{AgNO}_3\) to 0 in \(\mathrm{Ag}(s)\), gaining 1 electron.- Total electrons transferred is 2 (since 1 Cu loses 2 electrons and 2 Ag gain those 2 electrons).

Key Concepts

Oxidation Numbers

Oxidation numbers are vital in analyzing oxidation-reduction reactions, or "redox" reactions for short. They are used to track the transfer of electrons. Think of oxidation numbers as a sort of charge each atom has within a compound, which allows us to see how many electrons have been lost or gained.

In chemistry, oxidation numbers are often represented as integers. They help determine which atoms have been oxidized and which have been reduced in a reaction. For example, in the balanced reaction \( \text{Reaction (a)} \), elements like manganese (Mn) change from an oxidation state of \(+2\) to \(+7\). Here, manganese loses electrons, indicating it is oxidized.

Conversely, when we see bismuth (Bi) change from \(+5\) to \(+3\), it gains electrons, implying reduction. Each element in a compound is assessed, and the total oxidation number for a neutral compound adds up to zero. For example, oxygen will typically have an oxidation number of \(-2\). Once you get the hang of it, identifying these changes can become second nature.

Electron Transfer

Electron transfer is at the heart of oxidation-reduction reactions. It's the movement of electrons from one atom or molecule to another. This transfer is what essentially drives redox reactions, by allowing oxidizing and reducing agents to do their job.

Looking at \( \text{Reaction (b)} \), manganese (Mn) originally has an oxidation state of \(+7\) and is reduced to \(+4\) by gaining electrons. As manganese accepts electrons, it's effectively "closing the gap" that gives it a lower oxidation state.

Likewise, sulfur (S) in sodium sulfite \(\text{Na}_2\text{SO}_3 \) transforms from \(+4\) to \(+6\). In this case, electrons are lost, indicating oxidation. To determine the total electron transfer, one needs to ensure that the electrons lost and gained are balanced, maintaining the reaction's overall charge equilibrium. This is crucial in confirming the redox process's completeness.

Chemical Reactions

Chemical reactions transform substances through rearrangement of atoms, resulting in the formation of new products. Redox reactions, a specific type of chemical reaction, involve the exchange of electrons. This process alters the oxidation states of the participating elements.

Consider \( \text{Reaction (c)} \), where copper \((\text{Cu(s)})\) starts with an oxidation number of zero and ends with \(+2\) in \(\text{Cu(NO}_3)_2\). This transformation involves copper losing electrons and being oxidized. It's fascinating to see silver \((\text{Ag})\), initially silver nitrate \(\text{AgNO}_3\), do the reverse by gaining electrons and becoming elemental silver \((\text{Ag(s)}) \).

Understanding these reactions is essential in chemistry. Each atom's transfer and gain of electrons not only changes the compounds involved but also influences energy release or absorption, giving us insight into reaction energetics and making stoichiometry an indispensable tool in the study of chemistry.