Question

Consider the following unbalanced oxidation-reduction reactions in aqueous solution: $$\begin{array}{rl}{\mathrm{Ag}}^{+}(aq)+\mathrm{Li}(s)& ⟶\mathrm{Ag}(s)+{\mathrm{Li}}^{+}(aq)\\ \mathrm{Fe}(s)+{\mathrm{Na}}^{+}(aq)⟶{\mathrm{Fe}}^{2+}(aq)+\mathrm{Na}(s)\\ \mathrm{K}(s)+{\mathrm{H}}_2\mathrm{O}(l)& ⟶\mathrm{KOH}(aq)+{\mathrm{H}}_2(g)\end{array}$$ (a) Balance each of the reactions. (b) By using data in Appen$\mathrm{dix}C,$ calculate $\mathrm{\Delta }{H}^{\circ }$ for each of the reactions. (c) Based on the values you obtain for $\mathrm{\Delta }{H}^{\circ }$, which of the reactions would you expect to be thermodynamically favored? (d) Use the activity series to predict which of these reactions should occur. (Section 4.4) Are these results in accord with your conclusion in part (c) of this problem?

Short answer

(a) Balanced reactions: (1) ${\mathrm{Ag}}^{+}(aq)+\mathrm{Li}(s)\to \mathrm{Ag}(s)+{\mathrm{Li}}^{+}(aq)$ (2) $\mathrm{Fe}(s)+2{\mathrm{Na}}^{+}(aq)\to {\mathrm{Fe}}^{2+}(aq)+2\mathrm{Na}(s)$ (3) $\mathrm{K}(s)+{\mathrm{H}}_2\mathrm{O}(l)\to \mathrm{KOH}(aq)+\frac{1}{2}{\mathrm{H}}_2(g)$ (b) Use the given standard enthalpies of formation to calculate $\mathrm{\Delta }{H}_1^{\circ }$, $\mathrm{\Delta }{H}_2^{\circ }$, and $\mathrm{\Delta }{H}_3^{\circ }$ for each reaction. (c) The reaction with the lowest (most negative) ΔH° value is the most thermodynamically favored. (d) Compare the predictions from the activity series with the ΔH° values obtained in part (c) to check agreement.

Step-by-step solution

The first reaction

The half-reactions for the first reaction are: $${\mathrm{Ag}}^{+}\to \mathrm{Ag}$$ $$\mathrm{Li}\to {\mathrm{Li}}^{+}$$ Balanced reaction: ${\mathrm{Ag}}^{+}(aq)+\mathrm{Li}(s)\to \mathrm{Ag}(s)+{\mathrm{Li}}^{+}(aq)$

The second reaction

The half-reactions for the second reaction are: $$\mathrm{Fe}\to {\mathrm{Fe}}^{2+}$$ $${\mathrm{Na}}^{+}\to \mathrm{Na}$$ Balanced reaction: $\mathrm{Fe}(s)+2{\mathrm{Na}}^{+}(aq)\to {\mathrm{Fe}}^{2+}(aq)+2\mathrm{Na}(s)$

The third reaction

The half-reactions for the third reaction are: $$\mathrm{K}\to {\mathrm{K}}^{+}$$ $${\mathrm{H}}_2\mathrm{O}\to {\mathrm{OH}}^{-}+\frac{1}{2}{\mathrm{H}}_2$$ Balanced reaction: $\mathrm{K}(s)+{\mathrm{H}}_2\mathrm{O}(l)\to \mathrm{KOH}(aq)+\frac{1}{2}{\mathrm{H}}_2(g)$ Now, we move on to part (b) to calculate ΔH° for each of the reactions using data from Appendix C. (b) By using data in Appendix C, calculate ΔH° for each of the reactions.

Enthalpy Change Formula

We need the following formula for calculating the enthalpy change, ΔH°: $$\mathrm{\Delta }{H}^{\circ }=\sum \mathrm{\Delta }{H}_{\text{products}}^{\circ }-\sum \mathrm{\Delta }{H}_{\text{reactants}}^{\circ }$$

Calculating ΔH° for the first reaction

Let's assume the given data provides the standard enthalpy of formation for each species. Then, for the first reaction: $$\mathrm{\Delta }{H}_1^{\circ }=[(\mathrm{\Delta }{H}_{\text{Ag,s}}^{\circ })+(\mathrm{\Delta }{H}_{\text{Li^+,aq}}^{\circ })]-[(\mathrm{\Delta }{H}_{\text{Ag^+, aq}}^{\circ })+(\mathrm{\Delta }{H}_{\text{Li,s}}^{\circ })]$$

Calculating ΔH° for the second reaction

For the second reaction: $$\mathrm{\Delta }{H}_2^{\circ }=[(\mathrm{\Delta }{H}_{\text{Fe^{2+}, aq}}^{\circ })+(2\mathrm{\Delta }{H}_{\text{Na,s}}^{\circ })]-[(\mathrm{\Delta }{H}_{\text{Fe,s}}^{\circ })+(2\mathrm{\Delta }{H}_{\text{Na^+, aq}}^{\circ })]$$

Calculating ΔH° for the third reaction

For the third reaction: $$\mathrm{\Delta }{H}_3^{\circ }=[(\mathrm{\Delta }{H}_{\text{KOH,aq}}^{\circ })+(\frac{1}{2}\mathrm{\Delta }{H}_{\text{H\_2,g}}^{\circ })]-[(\mathrm{\Delta }{H}_{\text{K,s}}^{\circ })+(\mathrm{\Delta }{H}_{\text{H\_2O,l}}^{\circ })]$$ (c) Based on these values for ΔH°, we can determine which reaction is the most thermodynamically favored.

Thermodynamically favored reaction

Lower (more negative) values of ΔH° indicate a stronger tendency to release heat and favor the forward reaction. Therefore, among the calculated values, the reaction with the lowest ΔH° will be the most thermodynamically favored. (d) Use the activity series to predict which of these reactions should occur. Compare these results with the conclusion from part (c) of this problem.

Consulting the activity series

According to the activity series, we can predict whether a redox reaction will occur or not. In our case, we need to check the relative positions of the corresponding elements to determine which reaction will proceed spontaneously. Compare the predictions from the activity series with the ΔH° values obtained in part (c) to see if they are in agreement.

Key concepts

Enthalpy Change

In chemical reactions, exchanging energy with the surroundings is a common event, but quantifying this exchange is where the concept of enthalpy comes in. Enthalpy change, denoted as ΔH, measures the heat absorbed or released under constant pressure during a reaction. Calculating it isn't complicated, but you need specific data. The formula for determining the change in enthalpy is: $$\mathrm{\Delta }{H}^{\circ }=\sum \mathrm{\Delta }{H}_{\text{products}}^{\circ }-\sum \mathrm{\Delta }{H}_{\text{reactants}}^{\circ }$$ This calculation requires the standard enthalpies of formation for each compound involved, found in data tables like Appendix C. Each term considers the energy required or released when 1 mole of a substance is formed from its elements in their standard states.

• If ΔH is negative, the reaction releases energy to its surroundings, making it exothermic.
• If ΔH is positive, the reaction requires energy, making it endothermic.

Enthalpy changes give a clear picture of the energetic feasibility of reactions. Reactions that are strongly exothermic often proceed spontaneously, as they tend to move towards a state of lower energy.

Activity Series

The activity series is a crucial tool for predicting the potential and feasibility of oxidation-reduction reactions. It lists elements according to their relative reactivity, particularly how easily an element can lose or gain electrons. In this series, metals are generally ordered by their ability to displace H from acids or water. The higher an element is on the activity series, the more reactive it is and the more readily it will oxidize. This is essential for predicting whether a particular redox reaction can proceed.

• Elements higher in the series can displace those below them in compounds.
• For example, in the reaction K with water, K is high in the series and can displace hydrogen.

By comparing the reactants' positions, you can anticipate whether a transfer of electrons, and thus a reaction, will occur. Always remember, only metals higher in the series can replace metals lower in the series from solutions containing cations.

Thermodynamic Favorability

Thermodynamic favorability refers to the likelihood a reaction will occur under given conditions. It essentially combines insights from enthalpy change and entropy considerations. A reaction is considered thermodynamically favorable if it can proceed without any input of energy, often inferred from a negative Gibbs Free Energy change (ΔG). For enthalpy-driven processes:

• If ΔH is negative, the process tends to be more favorable.

But, favorability isn't solely about ΔH. You must also consider entropy, a measure of disorder in a system, and temperature. When ΔH and entropy changes are both favorable, a reaction will likely proceed. If they conflict, temperature often sways the actual path. Using thermodynamic principles helps delineate why and under what conditions reactions occur, ensuring deeper comprehension of chemical behavior.