Question

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ In this cell, the silver compartment contains a silver electrode and excess \(\operatorname{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)\), and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 M\). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. Assuming \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons & \\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned} $$

Short answer

The cell potential at \(25^{\circ} \mathrm{C}\) is approximately 0.61V. To obtain a cell potential of 0.52 V, approximately 1.556 moles of NH3 must be added.

Step-by-step solution

Write the Nernst equation

The Nernst equation relates the cell potential to the standard reduction potentials of the half-reactions and concentrations of species. The standard Nernst equation for an electrochemical cell is: $$E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln{Q}$$ Where: \(E_{cell}\) = cell potential (V) \(E^\circ_{cell}\) = standard cell potential (V) \(R\) = gas constant (8.314 J/molK) \(T\) = temperature (K) \(n\) = number of electrons transferred \(F\) = Faraday's constant (96485 C/mol) \(Q\) = reaction quotient

Find the standard cell potential

For Ag+ and Cu2+ half-reactions, the standard cell potentials are given: $$ \begin{aligned} \mathscr{E}^{\circ}(\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) ) = 0.80~\mathrm{V} \\ \mathscr{E}^{\circ}(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) ) = 0.34~\mathrm{V} \end{aligned} $$ Since silver has a higher standard cell potential, it serves as the cathode, and copper serves as the anode. Thus, theandard cell potential can be calculated as: $$E^\circ_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 0.80 - 0.34 = 0.46~\mathrm{V}$$

Calculate the reaction quotient Q

The overall cell reaction is: $$\mathrm{Cu}^{2+}(aq) + 2\mathrm{Ag}^{+}(aq) \rightarrow \mathrm{Cu}(s) + 2\mathrm{Ag}(s)$$ Since AgCl(s) is a solid, the reaction quotient \(Q\) is calculated as: $$Q = \frac{1}{[\mathrm{Cu}^{2+}]^1 [\mathrm{Ag}^{+}]^2} = \frac{1}{(2.0M \times K_{sp})^{\frac{1}{2}}}$$ Where \(K_{sp} = 1.6 \times 10^{-10}\) for AgCl.

Calculate the cell potential at 25°C

Substitute the values for \(E^\circ_{cell}\), \(Q\), \(n = 2\), and \(T=298K\) in the Nernst equation: $$E_{cell} = 0.46 - \frac{(8.314)(298)}{(2)(96485)} \ln{\left(\frac{1}{(2.0 \times 1.6 \times 10^{-10})^{\frac{1}{2}}}\right)} \approx 0.61~\mathrm{V}$$ Answer: The cell potential at \(25^{\circ} \mathrm{C}\) is approximately 0.61V. b. Assuming \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) .

Relate the given cell potential to the new E°cell

We are given a new cell potential of 0.52 V. Using the Nernst equation again: $$0.52 = E^\circ_{cell} - \frac{RT}{nF} \ln{Q_\text{new}}$$

Calculate Q_new

Solve for \(Q_\text{new}\): $$Q_\text{new} = e^{\left(\frac{0.52 - E^\circ_{cell}}{\frac{RT}{nF}}\right)} = e^{\left(\frac{0.52 - 0.46}{\frac{(8.314)(298)}{(2)(96485)}}\right)} \approx 3.895$$

Find the new concentration of Cu2+ ions

The new reaction quotient relates to the concentrations of the species: $$Q_\text{new} = \frac{1}{[\mathrm{Cu}^{2+}]_{\text{new}} [\mathrm{Ag}^{+}]^2} = \frac{1}{([\mathrm{Cu}^{2+}]_{\text{new}})(K_{sp})}.$$ After substituting the value for \(Q_\text{new}\) and solving for \([\mathrm{Cu}^{2+}]_{\text{new}}\), we get: $$[\mathrm{Cu}^{2+}]_{\text{new}} = \frac{1}{Q_\text{new} \times K_{sp}} \approx 1.611~\text{M}$$

Find the moles of NH3 added

Write the copper-ammine complex formation reaction and associate its equilibrium constant (\(K = 1.0 \times 10^{13}\)): $$\mathrm{Cu}^{2+}(aq) + 4\,\mathrm{NH}_{3}(aq) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(aq)$$ According to the reaction, 4 moles of NH3 react per mole of Cu2+ consumed. We are given initially 2.0 M Cu2+ in 1.0 L of solution, and the new concentration is approximately 1.611 M. Thus, moles of Cu2+ reacted are: \((2.0 - 1.611) \times 1.0 = 0.389~\text{mol}\) Since 4 moles of NH3 are required for every mole of Cu2+ reacted, the moles of NH3 added are: \(4 \times 0.389 = 1.556~\text{mol}.\) Answer: To obtain a cell potential of 0.52 V, approximately 1.556 moles of NH3 must be added.

Key concepts

Nernst Equation

The Nernst equation is a fundamental principle in electrochemistry, providing insight into the behavior of cells under non-standard conditions. It relates the measurable cell potential to the standard potential and the reaction quotient. The equation is given as:
\[E_{cell} = E^\text{o}_{cell} - \frac{RT}{nF} \ln Q\]
Here, \(E_{cell}\) represents the cell potential when the concentrations of reactants and products are not at standard conditions, while \(E^\text{o}_{cell}\) is the standard cell potential, indicating the potential difference when all components are in their standard states. The term 'standard state' typically means 1 molar concentration for solutions and 1 atmosphere pressure for gases at 25°C.
R is the universal gas constant (8.314 J/molK), T is the temperature in Kelvin, n is the number of moles of electrons transferred in the redox reaction, and F is the Faraday's constant, which represents the charge of one mole of electrons (96485 C/mol). Q is the reaction quotient, describing the ratio of the concentrations of products to reactants at any point during the reaction, not just at equilibrium.
The logarithmic term in the Nernst equation accounts for changes in cell potential due to concentration differences. As a cell operates, reactants are consumed, and products are formed, which affects the Q value and consequently the cell potential. Understanding the Nernst equation is crucial for predicting how a cell's potential will change over time as well as how varying concentrations will influence the voltage output of a cell.
For educational purposes, it's important to illustrate the Nernst equation with practical examples, similar to the one provided in the original exercise, to reinforce student understanding.

Reaction Quotient

The reaction quotient, Q, plays a pivotal role in assessing whether a reaction is at equilibrium and in determining the direction in which the reaction must proceed to reach equilibrium. For the generic chemical reaction \(aA + bB \rightarrow cC + dD\), the reaction quotient is expressed as:
\[Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
In this expression, the concentrations of the products are raised to the power of their stoichiometric coefficients and divided by the concentrations of the reactants raised to the power of their coefficients. Solids and pure liquids are omitted from the Q expression because their activities are considered to be constant. For example, in a galvanic cell reaction involving a solid like \(\mathrm{AgCl}(s)\), the concentration of \(\mathrm{AgCl}\) would not be included in Q because it does not change during the reaction.
The value of Q is compared with the equilibrium constant, K, to predict which direction the reaction must move to reach equilibrium. If \(Q < K\), the reaction proceeds forward; if \(Q > K\), the reaction proceeds in reverse; and if \(Q = K\), the reaction is at equilibrium. Understanding the role of Q in electrochemical cells allows us to predict and manipulate cell potentials as seen in the exercise with the addition of \(\mathrm{NH}_3\) to adjust the cell potential.

Standard Cell Potential

The standard cell potential, \(E^\text{o}_{cell}\), is the voltage or electrical potential difference between two half-cells when they are connected, assuming all reactants and products are at standard-state conditions. More specifically, it is the difference between the standard reduction potentials of the cathode and anode.
\[E^\text{o}_{cell} = E^\text{o}_{cathode} - E^\text{o}_{anode}\]
For instance, in the context of the exercise provided, we calculate the standard cell potential using the given potentials for the silver and copper half-reactions. Since the half-reaction with the higher standard potential tends to gain electrons (reduction), it serves as the cathode, while the one with the lower potential loses electrons (oxidation) and becomes the anode.
The importance of the standard cell potential lies in its ability to predict whether a cell can perform work spontaneously. A positive value for \(E^\text{o}_{cell}\) implies that the cell reaction is spontaneous under standard conditions, while a negative value suggests non-spontaneity. In electrochemistry, understanding standard cell potentials is essential for building batteries and other electrochemical devices, as well as for determining the feasibility of redox reactions.
To solidify student comprehension, we should emphasize the predictive nature of standard cell potentials in real-world applications, such as energy storage in batteries and corrosion prevention in metals, tying back to the original problem regarding galvanic cells.