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) $\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag (s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)$

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.

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).

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).

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).

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).

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).