Question

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V} .\) Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Short answer

In summary, for this electrochemical cell, we found that: a) The cell potential when \([\mathrm{Al^{3+}}] = 7.2 \times 10^{-3} \mathrm{M}\) is approximately \(1.68 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\). b) The concentration of \(\mathrm{Al^{3+}}\) in the unknown solution with a measured cell potential of \(1.62 \mathrm{V}\) at \(25^{\circ} \mathrm{C}\) is approximately \(1.01\text{ M}\).

Step-by-step solution

First, we need to write the balanced half-reactions for both nickel and aluminum electrodes in their reduction forms, and then combine these half-reactions to obtain the overall cell reaction. For nickel: \( \mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \leftrightarrow \mathrm{Ni}(s) \) For aluminum: \( \mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \leftrightarrow \mathrm{Al}(s) \) Now let's multiply half-reactions to have the same number of electrons, and then add them to find the overall cell reaction: \( 3(\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \leftrightarrow \mathrm{Ni}(s)) \) \( 2(\mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \leftrightarrow \mathrm{Al}(s)) \) Overall reaction: \( 3\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{Al}(s) \leftrightarrow 3\mathrm{Ni}(s) + 2\mathrm{Al}^{3+}(\mathrm{aq}) \) #Step 2: Determine the standard cell potential#

To calculate the standard cell potential (\(E^0_\mathrm{cell}\)), first, we need to find the standard reduction potentials for Ni²⁺ and Al³⁺: Standard reduction potential for Ni²⁺: \(\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Ni}(s)\): \(E^0_\mathrm{Ni} = -0.25 \mathrm{V}\) Standard reduction potential for Al³⁺: \(\mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}(s)\): \(E^0_\mathrm{Al} = -1.66 \mathrm{V}\) Now, we need to calculate the standard cell potential, which can be found using the following equation: \(E^0_\mathrm{cell} = E^0_\mathrm{cathode} - E^0_\mathrm{anode}\) Since aluminum is being oxidized, it will be the anode, and nickel will be the cathode: \(E^0_\mathrm{cell} = (-0.25 \mathrm{V}) - (-1.66 \mathrm{V}) = 1.41 \mathrm{V}\) #Step 3: Calculate the cell potential when [Al³⁺] = 7.2 x 10⁻³ M (part a)#

Now we need to use the Nernst equation to find the cell potential, which is: \(E_\mathrm{cell} = E^0_\mathrm{cell} - \frac{RT}{nF} \ln Q \) We are given the temperature (T = 25°C = 298.15 K) and can calculate the reaction quotient (Q). First, let's rewrite the overall reaction: \( 3\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{Al}(s) \leftrightarrow 3\mathrm{Ni}(s) + 2\mathrm{Al}^{3+}(\mathrm{aq}) \) The reaction quotient (Q) is given by: \(Q = \frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]^3}\) Plugging in the given concentrations: \(Q = \frac{(7.2 \times 10^{-3} \mathrm{M})^2}{(1.0 \mathrm{M})^3} = 5.184\times10^{-5} \) Now, we can calculate the cell potential. For this, we need the gas constant (R = 8.314 J mol⁻¹ K⁻¹), Faraday's constant (F = 96485 C mol⁻¹), and the number of moles of electrons transferred in the overall reaction (n = 6, since we need 6 electrons to balance the reaction): \(E_\mathrm{cell} = 1.41 \mathrm{V} - \frac{(8.314 \mathrm{J \cdot mol^{-1}\cdot K^{-1}})(298.15 \mathrm{K})}{(6)(96485\text{ C }\text{mol}^{-1})}\ln(5.184\times10^{-5})\) After calculating the cell potential: \(E_\mathrm{cell} = 1.68 \mathrm{V}\) (rounded to two decimal places) So, the potential of the cell at 25°C with [Al³⁺] = 7.2 x 10⁻³ M is 1.68 V. #Step 4: Calculate the concentration of Al³⁺ when the cell potential is 1.62 V (part b)#

We are given the cell potential (1.62 V) and need to find the corresponding concentration of Al³⁺. We can use the Nernst equation again: \(1.62 \mathrm{V} = 1.41 \mathrm{V} - \frac{(8.314 \mathrm{J \cdot mol^{-1}\cdot K^{-1}})(298.15 \mathrm{K})}{(6)(96485\text{ C }\text{mol}^{-1})}\ln Q \) Now, let's solve for Q: \(\ln Q = \frac{(1.62 \mathrm{V} - 1.41 \mathrm{V})(6)(96485\text{ C }\text{mol}^{-1})}{(8.314 \mathrm{J \cdot mol^{-1}\cdot K^{-1}})(298.15 \mathrm{K})}\) \(Q = e^{0.0141} \approx 1.014 \) Now we can solve for the concentration of Al³⁺: \(Q = \frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]^3} \Rightarrow [\mathrm{Al^{3+}}] = \sqrt{1.014 \times (1.0 \mathrm{M})^3}\) After calculating the Al³⁺ concentration: \([\mathrm{Al^{3+}}] \approx 1.01\text{ M} \) (rounded to two decimal places) So, in the unknown solution with a cell potential of 1.62 V, the concentration of Al³⁺ is approximately 1.01 M.

Key concepts

Understanding the Nernst Equation

The Nernst equation plays a central role in electrochemistry as it connects the standard cell potential, temperature, and the reaction quotient to the actual cell potential under non-standard conditions.

In essence, the Nernst equation adjusts the standard cell potential by accounting for the actual concentrations of reactants and products. It is mathematically expressed as: \[\begin{equation}E_{\text{cell}} = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \end{equation}\]where:

• \(E_{\text{cell}}\) is the actual cell potential,
• \(E^0_{\text{cell}}\) is the standard cell potential,
• \(R\) is the universal gas constant (8.314 J mol⁻¹ K⁻¹),
• \(T\) is the temperature in Kelvin,
• \(n\) is the number of moles of electrons transferred in the reaction,
• \(F\) is Faraday's constant (96485 C mol⁻¹),
• \(Q\) is the reaction quotient.

To apply it effectively, you need correct values for all these variables. The equation becomes extremely useful when dealing with electrochemical cells operating under non-standard conditions - that is, when the concentrations of reactants and products are different from their standard state values.

Standard Reduction Potentials

Standard reduction potentials are a fundamental concept in the study of electrochemical reactions. They essentially provide a measure of the inherent tendency of a species to gain electrons and reduce itself. The more positive the standard reduction potential, the greater the species' affinity for electrons.

• The standard reduction potential for nickel (\(\mathrm{Ni}^{2+}\)) is \(-0.25 V\), indicating a lesser tendency to gain electrons compared to aluminum (\(\mathrm{Al}^{3+}\)), whose standard reduction potential is significantly more negative at \(-1.66 V\).
• To determine the standard cell potential (\(E^0_{\text{cell}}\)), we subtract the standard potential at the anode from that at the cathode:\[\begin{equation}E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}}\end{equation}\]

These values are pivotal in calculations involving the Nernst equation as they provide the baseline or 'zero-point' from which actual potentials are measured. Remember this: a high-standard reduction potential means a strong pull for electrons, akin to a high capacity for reduction.

The Reaction Quotient (Q)

The reaction quotient (\(Q\)) is a measure that compares the relative amounts of products and reactants present during a reaction at a given moment. It is often used in the context of le Châtelier's principle and equilibrium, but it's equally important in electrochemistry when calculating cell potentials using the Nernst equation. \[\begin{equation}Q = \frac{[\text{products}]}{[\text{reactants}]}\end{equation}\]Although it looks similar to the equilibrium constant (\(K\)), \(Q\) can be applied to any state the reaction, not just at equilibrium. For the redox reaction described in our problem:\[\begin{equation}Q = \frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]^3}\end{equation}\]In using \(Q\) in the Nernst equation, we're effectively factoring in how the concentrations of reactants and products shift the cell potential away from its standard value. The reaction quotient tells us the immediate 'position' of the reaction - favoring products, favoring reactants, or at equilibrium - thus influencing the cell's electromotive force.