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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

Expert verified
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

01

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 Ni虏鈦 and Al鲁鈦: Standard reduction potential for Ni虏鈦: \(\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Ni}(s)\): \(E^0_\mathrm{Ni} = -0.25 \mathrm{V}\) Standard reduction potential for Al鲁鈦: \(\mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \rightarrow \mathrm{Al}(s)\): \(E^0_\mathrm{Al} = -1.66 \mathrm{V}\) 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 [Al鲁鈦篯 = 7.2 x 10鈦宦 M (part a)#
02

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掳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鈦宦 K鈦宦), Faraday's constant (F = 96485 C mol鈦宦), 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掳C with [Al鲁鈦篯 = 7.2 x 10鈦宦 M is 1.68 V. #Step 4: Calculate the concentration of Al鲁鈦 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 Al鲁鈦. 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 Al鲁鈦: \(Q = \frac{[\mathrm{Al^{3+}}]^2}{[\mathrm{Ni^{2+}}]^3} \Rightarrow [\mathrm{Al^{3+}}] = \sqrt{1.014 \times (1.0 \mathrm{M})^3}\) After calculating the Al鲁鈦 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 Al鲁鈦 is approximately 1.01 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈦宦 K鈦宦),
  • \(T\) 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鈦宦),
  • \(Q\) 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 Ch芒telier'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.

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Most popular questions from this chapter

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

It takes \(15 \mathrm{kWh}\) (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the HallHeroult process. Compare this to the amount of energy necessary to melt \(1.0 \mathrm{~kg}\) aluminum metal. Why is it economically feasible to recycle aluminum cans?

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 immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} M\) and the measured cell potential is \(1.82 \mathrm{~V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=?$$

When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q)\), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when \(\mathrm{Mg}\) is added directly to the beaker of HCl? How can you harness this reaction to do useful work?

A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?

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