Question

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10 \(M\) NaOH that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\mathrm{For} \mathrm{Cu}(\mathrm{OH})_{2}\right.\) \(K_{\mathrm{sp}}=1.6 \times 10^{-19} \cdot \mathrm{J}\)

Short answer

The cell potential at \(25^{\circ} \mathrm{C}\) for the given electrochemical cell consisting of a standard hydrogen electrode and a copper metal electrode in a solution of 0.10 M NaOH saturated with \(\mathrm{Cu}(\mathrm{OH})_{2}\) is approximately 0.402 V.

Step-by-step solution

Write the half-reactions

For the hydrogen electrode (standard conditions), the half-reaction is: \[ 2\mathrm{H}^{+}(aq) + 2e^{-} \rightarrow \mathrm{H}_{2}(g) \] For the copper metal electrode placed in a solution of 0.10 M NaOH that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2}\), the half-reaction is: \[ \mathrm{Cu}^{2+} + 2e^{-} \rightarrow \mathrm{Cu}(s) \] The overall cell reaction is the sum of these two half-reactions.

Determine the standard electrode potentials

The standard hydrogen electrode (SHE) is assigned a potential of 0 V by convention. The standard electrode potential of the copper half-reaction, \(E^0(\mathrm{Cu}^{2+}/\mathrm{Cu})\), is given as +0.34 V.

Write the solubility equilibrium expression for \(\mathrm{Cu}(\mathrm{OH})_{2}\)

For the equilibrium of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in water, we have: \[ \mathrm{Cu}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cu}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \] The solubility product constant expression can be written as: \[ K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^{2} \]

Calculate the concentration of \(\mathrm{Cu}^{2+}\) ions in the solution

Given the solubility product constant \(K_{sp} = 1.6 \times 10^{-19}\) and the concentration of \(\mathrm{OH}^{-}\) ions is 0.10 M (from 0.10 M NaOH), we can solve for the concentration of \(\mathrm{Cu}^{2+}\) ions: \[ 1.6 \times 10^{-19} = [\mathrm{Cu}^{2+}](0.10)^{2} \] Solving for \([\mathrm{Cu}^{2+}]\): \[ [\mathrm{Cu}^{2+}] = \frac{1.6 \times 10^{-19}}{(0.10)^{2}} = 1.6 \times 10^{-17} \mathrm{M} \]

Calculate the cell potential at \(25^{\circ} \mathrm{C}\) using the Nernst equation

The Nernst equation can be used to determine the cell potential at non-standard conditions: \[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \] Where \(E_{cell}\) is the cell potential at the given conditions, \(E^0_{cell}\) is the standard cell potential, R is the gas constant (8.314 J/(mol K)), T is the temperature in Kelvin (\(25^{\circ} \mathrm{C}\) = 298.15 K), n is the number of electrons transferred in the reaction (2 in this case), F is the Faraday's constant (96,485 C/mol), and Q is the reaction quotient. First, calculate the standard cell potential: \[ E^0_{cell} = E^0(\mathrm{Cu}^{2+}/\mathrm{Cu}) - E^0(\mathrm{H}^{+}/\mathrm{H}_{2}) = 0.34 \mathrm{V} - 0 \mathrm{V} = 0.34 \mathrm{V} \] Next, calculate the reaction quotient Q as \[ Q = \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{H}^{+}]} \] Since the pH of the half-reaction with the hydrogen electrode is 7 (neutral), the concentration of \(\mathrm{H}^{+}\) ions is \(1 \times 10^{-7}\) M. Therefore, Q becomes: \[ Q = \frac{1.6 \times 10^{-17}}{1 \times 10^{-7}} = 1.6 \times 10^{-10} \] Now plug the values into the Nernst equation: \[ E_{cell} = 0.34 - \frac{8.314 \times 298.15}{2 \times 96485} \ln (1.6 \times 10^{-10}) \] Solve for \(E_{cell}\): \[ E_{cell} \approx 0.402 \mathrm{V} \] So, the cell potential at \(25^{\circ} \mathrm{C}\) for the given electrochemical cell is approximately 0.402 V.

Key concepts

Nernst Equation

The Nernst equation is a fundamental formula used to determine the cell potential at any given concentration of reactants and products, not just under standard conditions. It introduces a way to account for the changes in potential based on different ion concentrations.

The general form of the Nernst equation is:
\[E = E^0 - \frac{RT}{nF} \ln Q\]
where \(E\) is the cell potential under non-standard conditions, \(E^0\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the electrochemical reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient, which represents the ratio of the concentrations of the products raised to the power of their stoichiometric coefficients to the reactants raised to the power of their coefficients in the balanced equation.

For example, when calculating the cell potential using different concentrations of ions than at standard conditions, the Nernst equation adjusts the cell voltage to reflect these changes. This is especially useful in galvanic cells where the concentration of ions can easily change due to the reactions occuring at the electrodes.

Standard Electrode Potentials

Standard electrode potentials, denoted by \(E^0\), are the electrical potentials of half-reactions measured under standard conditions: 1M concentration of ions, a pressure of 1 atmosphere for gases, and a temperature of 298 K (25°C). Each half-reaction has a standard electrode potential, which, when combined properly, allow us to calculate the overall standard cell potential for full reactions in electrochemical cells.

The most common reference point used for these measurements is the standard hydrogen electrode (SHE), which is assigned a potential of 0 V. The standard electrode potential can tell us the tendency of a substance to gain or lose electrons—also known as its reducing or oxidizing power. When we have two half-cells under standard conditions, the standard cell potential \(E^0_{cell}\) is the difference between the potential of the cathode and the anode.

The standard electrode potential for the copper electrode in the provided exercise is +0.34 V, and when compared to the standard hydrogen electrode, which is 0 V, it helps us calculate the overall standard cell potential.

Solubility Product Constant (Ksp)

The solubility product constant, represented by \(K_{sp}\), is a special type of equilibrium constant that measures the solubility of a solid in an aqueous solution. It's particularly important for slightly soluble salts. It takes into account the maximum amount of the salt that can dissolve in water to form a saturated solution.

For the dissolution of a generic salt \(AB\) in water, the reaction can be written as:
\[AB(s) \leftrightarrows A^{+}(aq) + B^{-}(aq)\]
The solubility product constant for this reaction would be:
\[K_{sp} = [A^{+}][B^{-}]\]
where the concentrations are those at equilibrium in a saturated solution. In the problem provided, we calculate the concentration of \(Cu^{2+}\) ions in the solution using the solubility product constant. For substances with low solubility, knowing \(K_{sp}\) can help predict whether a precipitate will form under certain conditions and is key for understanding the solubility behavior of compounds in different environments.

Half-reactions

Electrochemical processes often involve redox reactions, which can be broken down into two half-reactions: oxidation (loss of electrons) and reduction (gain of electrons). These half-reactions are an essential concept for understanding how electrochemical cells operate.

Each half-reaction occurs at a different electrode in an electrochemical cell: the anode for oxidation and the cathode for reduction. By separating the overall reaction into half-reactions, we simplify the study of the processes and make it easier to balance and understand the reactions occurring within the cell. Moreover, writing half-reactions enables us to assign standard electrode potentials to each half-reaction, which are fundamental to the calculation of the cell's electromotive force (emf).

In the given exercise, the half-reaction at the hydrogen electrode is the standard: \(2H^{+}(aq) + 2e^{-} \rightarrow H_2(g)\), with 0 V potential. On the other hand, the copper half-reaction is \(Cu^{2+} + 2e^{-} \rightarrow Cu(s)\), with a potential of +0.34 V. These two half-reactions help us understand the overall redox process and the movement of electrons that generate electrical energy within the cell.