Question

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+ \mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(g)+ 5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)\)

Short answer

For the given reactions, the equilibrium constants at 298 K are: (a) \(K \approx 1.26 \times 10^{13}\) (b) Use the same process to find the equilibrium constant for reaction (b). (c) Use the same process to find the equilibrium constant for reaction (c).

Step-by-step solution

Stage 1: Determine the net standard reduction potential (E)

To calculate the net standard reduction potential for the reaction, we must first identify the half-reactions: Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\) Reduction: \(2 \mathrm{Ag}^{+}(a q) + 2e^- \longrightarrow 2 \mathrm{Ag}(s)\) Next, look up the standard reduction potentials in Appendix E: \(E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = +0.34 \mathrm{V}\) (reverse the sign since it's an oxidation) \(E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} = +0.80 \mathrm{V}\) Now, calculate the net standard reduction potential: \(E = E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} + E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} = -0.34 \mathrm{V} + 0.80 \mathrm{V} = 0.46 \mathrm{V}\)

Stage 2: Calculate the change in Gibbs free energy (\(\Delta G\))

Use the relationship between Gibbs free energy and the standard reduction potentials: \(\Delta G = -nFE = -2 \times 96485 \mathrm{C/mol} \times 0.46 \mathrm{V} = -88685.8 \mathrm{J/mol}\)

Stage 3: Calculate the equilibrium constant (K)

Using the formula for equilibrium constant: \(K = 10^{\frac{-\Delta G}{RT}} = 10^{\frac{88685.8 \mathrm{J/mol}}{8.314 \mathrm{J/mol}\cdot\mathrm{K} \times 298 \mathrm{K}}} = 10^{13.1} \approx 1.26 \times 10^{13}\) The equilibrium constant for reaction (a) is \(K \approx 1.26 \times 10^{13}\). Repeat the same process for reactions (b) and (c) to find their respective equilibrium constants.

Key concepts

Standard Reduction Potential

In redox reactions, the standard reduction potential (denoted as \(E^\circ\)) is a crucial concept. It refers to the tendency of a chemical species to gain electrons and be reduced.
Standard reduction potentials are measured under specific conditions: 1 molar solution concentration, a pressure of 1 atm for any gases involved, and a temperature of 298 K (25°C).
This measurement is relative to the standard hydrogen electrode, which has a defined potential of 0 V.
These potentials are listed in tables, such as Appendix E in many chemistry textbooks.

To assess a redox reaction, we break it down into two half-reactions: oxidation and reduction. Each half-reaction's potential is looked up in a table.
The net standard reduction potential is determined by subtracting the oxidation potential from the reduction potential.
For instance, if the oxidation potential of copper is -0.34 V (since it's actually given for reduction, you reverse its sign for oxidation), and the reduction potential for silver is +0.80 V, the net potential \(E\) is calculated as follows:
\[ E = (-0.34 \text{ V}) + (0.80 \text{ V}) = 0.46 \text{ V} \].
This resulting potential helps predict the spontaneity of a reaction.

Gibbs Free Energy

Gibbs free energy (\(\Delta G\)) provides insights into the thermodynamic feasibility of a chemical reaction.
It tells us whether a reaction can occur spontaneously at constant temperature and pressure.
The link between Gibbs free energy and the equilibrium state of a reaction is given by the equation:
\[ \Delta G = -nFE \]
where \(n\) is the number of moles of electrons transferred, \(F\) is Faraday's constant (approximately 96485 C/mol), and \(E\) is the net standard reduction potential.
If \(\Delta G\) is negative, the process is spontaneous.
For example, for a redox reaction transferring two moles of electrons with a net standard reduction potential of 0.46 V, \(\Delta G\) is:

\[ \Delta G = -2 \times 96485 \text{ C/mol} \times 0.46 \text{ V} = -88685.8 \text{ J/mol} \]

This negative value confirms the reaction will proceed without additional energy.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between species.
One substance undergoes oxidation (loses electrons), while another undergoes reduction (gains electrons).
Key to understanding redox reactions is recognizing these two simultaneous processes.
For any given redox reaction, you should identify the substances being oxidized and reduced.
In the example reaction \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\), copper loses electrons to become \(\mathrm{Cu}^{2+}\) (oxidation), and silver ions gain electrons to form solid silver (reduction).
Balancing redox reactions involves ensuring the conservation of charge and atoms.
For complex reactions, breaking them into smaller half-reactions often clarifies the electron transfer dynamics.
Learning to quickly balance and interpret these reactions lays the groundwork for calculating equilibrium constants and understanding chemical spontaneity.

Electrochemical Cell

An electrochemical cell is a device that generates electricity through redox reactions.
There are mainly two types: galvanic (or voltaic) cells, which produce electrical energy through spontaneous redox reactions, and electrolytic cells, which consume electrical energy to drive non-spontaneous reactions.
In a galvanic cell, the reduction and oxidation half-reactions occur in separate compartments, known as half-cells.
The electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs) via an external circuit.
Often, a salt bridge completes the circuit, allowing ions to move and maintain neutrality.
Electrochemical cells help us understand energy conversion, corrosion processes, and even the functioning of batteries.
They illustrate how chemical energy is transformed into electrical energy efficiently. Understanding them helps students connect theoretical knowledge of redox reactions with practical applications, such as powering devices or understanding the chemistry behind batteries.