Question

A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C}\) : $$\begin{aligned}\mathrm{Ag}^{+}+\mathrm{e}^{-} & \longrightarrow \mathrm{Ag} \\\ \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ Predict whether \(\mathscr{G}_{\text {cell }}\) is larger or smaller than \(\mathscr{C}_{\text {cell }}^{\circ}\) for the following cases. a. \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\)

Short answer

In summary, for the given cases a and b, the relationship between ΔG and ΔG° is as follows: a. ΔG = ΔG° (because Q = 1) b. ΔG < ΔG° (because Q > 1)

Step-by-step solution

Balancing the overall cell reaction

First, we need to balance the given half-reactions. We notice that for the two reactions to be balanced, we need to multiply the first half-reaction by 2 so that there are equal numbers of electrons: 1. 2(Ag⁺ + e⁻ → Ag) 2. H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O Now, since the number of electrons on both sides is equal, we can add the half-reactions: 2Ag⁺ + H₂O₂ + 2H⁺ → 2Ag + 2H₂O This is the balanced overall cell reaction. Step 2: Calculate the reaction quotient, Q

Reaction Quotient

Using the balanced overall cell reaction, let's determine the reaction quotient (Q): Q = \(\frac{[Ag]^2[H_2O]^2}{[Ag^+]^2[H_2O_2][H^+]^2}\) Step 3: Determine the relationship between ΔG and ΔG°

Case a: Concentrations Ag⁺ = 1.0 M, H₂O₂ = 2.0 M, H⁺ = 2.0 M

Using the given concentrations, let's calculate the reaction quotient Q for case a: Q = \(\frac{[Ag]^2[H_2O]^2}{[Ag^+]^2[H_2O_2][H^+]^2}\) = \(\frac{[1]^2[2]^2}{[1]^2[2][2]^2}\) = 1 Since Q = 1, we can infer that ΔG = ΔG°. Therefore, in case a, ΔG is equal to ΔG°.

Case b: Concentrations Ag⁺ = 2.0 M, H₂O₂ = 1.0 M, H⁺ = 1.0 x 10⁻⁷ M

Using the given concentrations, let's calculate the reaction quotient Q for case b: Q = \(\frac{[Ag]^2[H_2O]^2}{[Ag^+]^2[H_2O_2][H^+]^2}\) = \(\frac{[2]^2[1]^2}{[2]^2[1][1.0 \times 10^{-7}]^2}\) Q = 1 / (1.0 × 10⁻¹⁴) = 1.0 × 10¹⁴ Since Q is greater than 1, we know ΔG < ΔG°. Therefore, in case b, ΔG is smaller than ΔG°.

Key concepts

Galvanic cells

Galvanic cells, also known as voltaic cells, are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two different electrodes immersed in electrolyte solutions, which are connected by a salt bridge or porous disk that allows ions to move between the solutions. This flow of ions maintains electrical neutrality.

The galvanic cell operates by separating the oxidation and reduction reactions into two half-cells, each containing a respective electrode and solution. The anode is where oxidation occurs, losing electrons, and the cathode is where reduction takes place, gaining electrons. Electrons flow from the anode to the cathode through an external circuit, generating an electric current.

In the problem, the half-reactions involve the reduction of silver ions to silver (Ag) at one electrode, and the reduction of hydrogen peroxide to water at another. By understanding the currents in these reactions, we can determine the direction and magnitude of the potential generated by the cell.

Reaction quotient

In electrochemistry, the reaction quotient, represented by the symbol \(Q\), is a measure of the relative concentrations of products and reactants at a given moment in time. It is similar to the equilibrium constant \(K\), but \(Q\) can be calculated for any set of conditions, not just at equilibrium.

For a given reaction, the expression for \(Q\) involves the concentration of the products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their coefficients:

• \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

In the exercise, for the reaction:

• \(2\mathrm{Ag}^+ + \mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ \rightarrow 2\mathrm{Ag} + 2\mathrm{H}_2\mathrm{O}\)

The reaction quotient \(Q\) is given by:

• \(Q = \frac{[\mathrm{Ag}]^2 [\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{Ag}^+]^2 [\mathrm{H}_2\mathrm{O}_2] [\mathrm{H}^+]^2}\)

Understanding \(Q\) helps predict the direction of the reaction. If \(Q < K\), the forward reaction is favored. If \(Q > K\), the reverse reaction occurs. And if \(Q = K\), the system is at equilibrium.

Thermodynamics

Thermodynamics in the context of electrochemistry is about the energy changes that occur during redox reactions. A critical part of this is the Gibbs free energy change \(\Delta G\), which indicates whether a reaction is spontaneous. In a galvanic cell, \(\Delta G\) is related to the cell potential (voltage) \(E\) by the equation:

• \(\Delta G = -nFE\)

where \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday’s constant.

The standard free energy change \(\Delta G^\circ\) applies under standard conditions (1 M concentration, 1 atm pressure, 25°C). However, reactions often occur at non-standard conditions, so we use the relationship:

• \(\Delta G = \Delta G^\circ + RT\ln(Q)\)

where \(R\) is the gas constant and \(T\) is the temperature in Kelvin.

By calculating \(Q\) and determining how \(\Delta G\) compares to \(\Delta G^\circ\), we can predict if a cell reaction will proceed and potentially generate energy. In the exercise, different conditions show how \(Q\) affects \(\Delta G\), helping us understand the behavior of the cells.

Half-reactions

Half-reactions are a way to separate the oxidation and reduction processes of a redox reaction. Each half-reaction shows either the loss of electrons (oxidation) or the gain of electrons (reduction). In a galvanic cell, these two half-reactions take place in separate compartments, or half-cells.

The purpose of writing and balancing half-reactions is to show clearly how electrons are transferred between species. This is essential for understanding the flow of electricity in an electrochemical cell. For the given exercise:

• The reduction half-reaction for silver ions is: \(\mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag}\)
• The reduction half-reaction for hydrogen peroxide is: \(\mathrm{H}_2\mathrm{O}_2 + 2\mathrm{H}^+ + 2\mathrm{e}^- \rightarrow 2\mathrm{H}_2\mathrm{O}\)

Once both half-reactions are balanced for charge and mass, they can be combined to give the overall cell reaction. This balanced reaction helps us calculate cell potentials and evaluate how the cell performs under different conditions, crucial for determining its efficiency.