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) \(14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)\) \(\quad \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q)\) (b) \(2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(\quad \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q)\) (c) \(\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag (s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)\)

Short answer

In summary: (i) Oxidation numbers in each reaction are: (a) Reactants: H(+1), Mn(+2), Na(+1), Bi(+3), O(-2); Products: H(+1), Mn(+7), O(-2), Bi(+3), Na(+1). (b) Reactants: K(+1), Mn(+7), O(-2), Na(+1), S(+4), H(+1); Products: Mn(+4), O(-2), Na(+1), S(+6), K(+1), H(+1). (c) Reactants: Cu(0), Ag(+1), N(+5), O(-2); Products: Ag(0), Cu(+2), N(+5), O(-2). (ii) Electrons transferred in each reaction: (a) 10 electrons (b) 6 electrons (c) 2 electrons

Step-by-step solution

Reaction (a) Oxidation Numbers

To find the oxidation numbers of the elements in the reactants and products, we apply the oxidation number rules in the order listed. For 14H+(aq): - H has an oxidation number of +1, since it's a hydrogen ion. For 2Mn^2+(aq): - Mn has an oxidation number of +2, since it's a manganese ion. For 5NaBiO3(s): - Na has an oxidation number of +1, as alkali metals usually have an oxidation number of +1. - Bi has an oxidation number of +3, as group 15 elements typically have an oxidation number of +3 in their compounds. - O has an oxidation number of -2, since it's an oxide ion. For 7H2O(l): - O has an oxidation number of -2, as oxygen in compounds (except peroxides) typically has an oxidation number of -2. - H has an oxidation number of +1, as hydrogen typically has an oxidation number of +1 in compounds. For 2MnO4^-(aq): - Mn has an oxidation number of +7, as the sum of the oxidation numbers of elements in a compound should equal the charge of the compound (-1 in this case). So, (+7) + 4(-2) = -1. - O has an oxidation number of -2 (as before). For 5Bi^3+(aq): - Bi has an oxidation number of +3, since it's a bismuth ion. For 5Na+(aq): - Na has an oxidation number of +1, as alkali metals usually have an oxidation number of +1.

Reaction (a) Electrons Transferred

To find the total number of electrons transferred in the reaction, we analyze the change in oxidation numbers of each element. In this case, we're focusing on Mn and Bi. Mn goes from +2 to +7, meaning it loses 5 electrons per atom. There are two Mn atoms, so it loses a total of 10 electrons in the process. Bi goes from +3 to +3; it has no change in oxidation number, meaning no electrons were transferred for Bi. Thus, there were 10 electrons transferred in reaction (a).

Reaction (b) Oxidation Numbers

Apply the oxidation number rules as mentioned before. For 2KMnO4(aq), Na2SO3(aq), and H2O(l), we have the same oxidation numbers for K, Mn, O, Na, S, and H as in reaction (a). As for the products: For 2MnO2(s): - Mn has an oxidation number of +4, as the sum of the oxidation numbers should equal zero. So, (+4) + 2(-2) = 0. - O has an oxidation number of -2 (as before). For 3Na2SO4(aq): - Na, S, and O have the same oxidation numbers as in Na2SO3(aq). For 2KOH(aq): - K has an oxidation number of +1 (as before). - O has an oxidation number of -2 (as before). - H has an oxidation number of +1 (as before).

Reaction (b) Electrons Transferred

The elements that undergo oxidation/reduction are Mn and S. Mn goes from +7 to +4, meaning it gains 3 electrons per atom. There are two Mn atoms, so it gains a total of 6 electrons in the process. S goes from +4 to +6, meaning it loses 2 electrons per atom. There are three S atoms, so it loses a total of 6 electrons in the process. Thus, there were 6 electrons transferred in reaction (b).

Reaction (c) Oxidation Numbers

We apply the rules as before. For Cu(s): - Cu has an oxidation number of 0, since elements in their elemental state have an oxidation number of 0. For 2AgNO3(aq): - Ag has an oxidation number of +1, since it's a silver ion. - N has an oxidation number of +5, as the sum of the oxidation numbers should equal zero in the compound nitrate. So, (+5) + 3(-2) = 0. - O has an oxidation number of -2 (as before). For 2Ag(s): - Ag has an oxidation number of 0, since elements in their elemental state have an oxidation number of 0. For Cu(NO3)2(aq): - Cu has an oxidation number of +2, since it's a copper(II) ion. - N and O have the same oxidation numbers as in AgNO3(aq).

Reaction (c) Electrons Transferred

The elements that undergo oxidation/reduction are Cu and Ag. Cu goes from 0 to +2, meaning it loses 2 electrons per atom. There is one Cu atom, so it loses a total of 2 electrons in the process. Ag goes from +1 to 0, meaning it gains 1 electron per atom. There are two Ag atoms, so it gains a total of 2 electrons in the process. Thus, there were 2 electrons transferred in reaction (c).

Key concepts

Oxidation Numbers

An oxidation number is a figure that represents the total number of electrons an atom uses to bond with other atoms. This concept is fundamental in understanding how atoms interact in oxidation-reduction (redox) reactions. Assigning oxidation numbers helps us track which atoms lose or gain electrons during a chemical reaction.

Here are some key rules for determining oxidation numbers:

• The oxidation number of an atom in its elemental form is always 0, as seen with solid copper (Cu) and monatomic silver (Ag) in Exercise (c).
• For ions consisting of a single atom (monatomic ions), the oxidation number is the same as the ion's charge. For example, in Exercise (a), the oxidation number of Na in NaBiO₃ is +1.
• Oxygen typically has an oxidation number of -2, except in peroxides or when bonded with fluorine.
• Hydrogen is usually +1 when bonded with non-metals, aligning with its oxidation number in water (H₂O) in Exercise (a).
• For other elements, the oxidation number is determined so that the sum of oxidation numbers for a compound matches its overall charge. For instance, Mn in MnO₄^- in Exercise (a) is +7 to balance the anion's -1 charge.

Recognizing these numbers is important, as you can see changes that signal a redox reaction, such as the shift from +2 to zero in Ag in Exercise (c), indicating a gain of electrons.

Electron Transfer

Electron transfer is the movement of electrons from one atom or molecule to another. It is the core process in oxidation-reduction reactions, as these involve either the loss (oxidation) or gain (reduction) of electrons.

To count the number of electrons transferred in a reaction, follow these steps:

• Identify the atom undergoing oxidation and the one undergoing reduction.
• Determine the change in their oxidation numbers. For instance, in Exercise (a), Mn's oxidation state changes from +2 to +7, signifying a loss of 5 electrons.
• Multiply the number of atoms by the difference in their oxidation states to find the total number of electrons transferred. Two Mn atoms losing 5 electrons each results in a 10-electron shift in reaction (a).

This transfer is pivotal, as it tells us how many electrons move overall, providing insight into the electron balance and stoichiometry necessary to keep reactions balanced and accurate.

Chemical Reactions

Chemical reactions are processes where substances called reactants transform into new substances known as products. Oxidation-reduction reactions, or redox reactions, is a category of chemical reactions involving the transfer of electrons between different substances.

In a redox reaction:

• The substance that loses electrons is said to be oxidized. This occurs when there's an increase in the oxidation number, such as copper (Cu) in Exercise (c) moving from 0 to +2, losing electrons and being oxidized.
• The substance that gains electrons is reduced, shown by a decrease in its oxidation number. For example, silver (Ag) in Exercise (c) decreases from +1 to 0 as it gains an electron and is reduced.
• Both oxidation and reduction must occur together, ensuring conservation of mass and charge in chemical processes. A real-world analogy would be a seesaw, where one side lowers (reduction) as the other rises (oxidation), maintaining balance.

Understanding these reactions is key in applications like electrochemistry, where redox principles enable the creation of batteries, fuel cells, and metabolic processes in biological systems.