Question

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{aligned} \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{Cu}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow & 2 \mathrm{I}^{-}(a q) \end{aligned} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=3.5 \mathrm{M} .\) (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) was equal to \(0.15 \mathrm{M},\) at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

Short answer

The answers to each part of the question are as follows: (a) The cell potential at the given concentrations is 0.045 V. (b) The anode of the cell is the electrode where the Cu+/Cu half-cell reaction occurs. (c) Yes, the answer to part (b) is the same as if the cell were operated under standard conditions. (d) If \([\mathrm{Cu}^{+}]\) is equal to \(0.15 \,\mathrm{M}\), the concentration of \(\mathrm{I}^-\) at which the cell has zero potential is \(0.165\, \mathrm{M}\).

Step-by-step solution

Determine the cell potential (E) at given concentrations

For this, we will use the Nernst equation: \[E = E^0 - \frac{RT}{nF} \ln Q\] where E is the cell potential, E0 is the standard cell potential, R is the gas constant (8.314 J/mol·K) , T is the temperature (298 K), n is the number of electrons transferred in the cell reaction, F is the Faraday constant (96485 C/mol) and Q is the reaction quotient. First, we need to find the standard cell potential (E0) using the standard reduction potentials: $$ \begin{aligned} E_{\mathrm{Cu}^{+} / \mathrm{Cu}}^0 &= +0.522\,\mathrm{V} \\ E_{\mathrm{I}_{2} / \mathrm{I}^-}^0 &= +0.536\,\mathrm{V} \end{aligned} $$ As we want the cell to have a positive cell potential, the cathode reaction should have a higher standard reduction potential, so the second reaction will be the cathode reaction, and the first one will be the anode reaction. So, $$ \begin{aligned} E^0 &= E_{\mathrm{cathode}}^0 - E_{\mathrm{anode}}^0 \\ &= (0.536 - 0.522) \,\mathrm{V} \\ &= 0.014\, \mathrm{V} \end{aligned} $$ The reaction quotient (Q) is given by: $$ Q = \frac{ [\mathrm{Cu}^{+}] }{ [\mathrm{I}^-]^2 } $$ Now, substituting the values for R, T, n, F, E0, and Q in the Nernst equation, we get: $$ \begin{aligned} E &= 0.014\,\mathrm{V} - \frac{8.314\, \mathrm{J} \cdot\mathrm{mol}^{-1} \cdot\mathrm{K}^{-1} \cdot 298\,\mathrm{K}}{2 \cdot 96485\,\mathrm{C}\cdot\mathrm{mol}^{-1}} \ln \left(\frac{0.25}{3.5^2}\right) \\ &= 0.014\,\mathrm{V} - 0.0042\,\mathrm{V} \ln \left(\frac{0.25}{3.5^2}\right) \\ E &= 0.045\, \mathrm{V} \end{aligned} $$ (a) The cell potential at these concentrations is 0.045 V.

Identify the anode of the cell

As we established earlier, the first reaction (Cu+ + e− -> Cu) is the anode reaction because it has a lower standard reduction potential. Therefore, the electrode where this reaction occurs is the anode of the cell. (b) The anode of the cell is the electrode where the Cu+/Cu half-cell reaction occurs.

Compare anode under standard conditions

Under standard conditions, the concentration of Cu+ and I- would be 1 M. Here, the standard half-cell potentials will be E0, and the reaction quotient is 1, so: $$ \begin{aligned} E &= 0.014\,\mathrm{V} - \frac{8.314\,\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \cdot 298\, \mathrm{K}}{2 \cdot 96485\, \mathrm{C}\cdot\mathrm{mol}^{-1}} \ln(1)\\ E &= 0.014\,\mathrm{V} \end{aligned} $$ The cell potential is still positive under standard conditions, so the same reaction will still be the anode reaction, and thus the anode remains the same. (c) Yes, the answer to part (b) is the same as it would be if the cell were operated under standard conditions.

Find the concentration of I- when cell potential is zero

If E = 0, then we will use the Nernst equation to find the concentration of I-: $$ \begin{aligned} 0 &= 0.014\,\mathrm{V} - \frac{8.314\, \mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \cdot 298\,\mathrm{K}}{2 \cdot 96485\,\mathrm{C}\cdot\mathrm{mol}^{-1}} \ln \left(\frac{0.15}{x^2}\right) \end{aligned} $$ To solve for x (concentration of I-), we first rearrange the equation to get the expression inside the logarithm: $$ \begin{aligned} \frac{8.314\, \mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \cdot 298\,\mathrm{K}}{2 \cdot 96485\,\mathrm{C}\cdot\mathrm{mol}^{-1}} \ln \left(\frac{0.15}{x^2}\right) &= 0.014\, \mathrm{V}\\ \ln \left(\frac{0.15}{x^2}\right) &= \frac{0.014\, \mathrm{V} \cdot 2 \cdot 96485\, \mathrm{C}\cdot\mathrm{mol}^{-1}}{8.314\, \mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \cdot 298\, \mathrm{K}} \\ \frac{0.15}{x^2} &= \exp\left(\frac{0.014\, \mathrm{V} \cdot 2 \cdot 96485\, \mathrm{C}\cdot\mathrm{mol}^{-1}}{8.314\, \mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \cdot 298\, \mathrm{K}}\right) \\ x &= \sqrt{\frac{0.15}{\exp\left(\frac{0.014\,\mathrm{V} \cdot 2 \cdot 96485\, \mathrm{C}\cdot\mathrm{mol}^{-1}}{8.314\, \mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1} \cdot 298\,\mathrm{K}}\right)}} \\ x &= 0.165\, \mathrm{M} \end{aligned} $$ (d) If \([\mathrm{Cu}^{+}]\) is equal to \(0.15 \,\mathrm{M}\), the concentration of \(\mathrm{I}^-\) at which the cell has zero potential is \(0.165\, \mathrm{M}\).

Key concepts

Nernst Equation

The Nernst equation is fundamental to electrochemistry and plays a crucial role in determining the cell potential (\(E\)) of an electrochemical cell under non-standard conditions. It establishes a relationship between the standard cell potential (\(E^0\)), temperature (\(T\)), the number of moles of electrons transferred in the reaction (\(n\)), the charge on a mole of electrons (Faraday's constant, \(F\)), and the reaction quotient (\(Q\)).

The equation is given by: \[E = E^0 - \frac{RT}{nF} \ln Q\] where \(E^0\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. The natural logarithm of the reaction quotient (\(\ln Q\) reflects the ratio of the concentrations of the reaction’s products to its reactants. The Nernst equation is a valuable tool because it allows us to calculate the potential of a cell at any combination of reactant and product concentrations, giving us insight into the cell's behavior in real-world applications.

Cell Potential

The cell potential, often referred to as the electromotive force (emf) of a voltaic cell, is a measure of the energy difference that drives the flow of electrons through the external circuit from the anode to the cathode. It is denoted by \(E\) and measured in volts (\(V\)).

The cell potential can be calculated under standard conditions, where all reactants and products are at 1 M concentration and at a temperature of 298 K. However, in practical situations, reactant and product concentrations can vary, altering the cell potential from its standard value. This is where applying the Nernst equation becomes essential to determine the actual cell potential. The standard cell potential (\(E^0\)) is found using standard reduction potentials of the respective half-cell reactions, by subtracting the standard reduction potential of the anode from the cathode.

Standard Reduction Potential

Standard reduction potentials (\(E^0\)) are used to determine the direction in which an electrochemical reaction is spontaneous under standard conditions. These potentials are measured in volts and can be found in tables of standard electrochemical potentials. A higher positive value indicates a greater tendency for the substance to gain electrons and be reduced.

In the example given, the standard reduction potentials for the copper and iodine half-cells were provided which allowed us to calculate the cell's standard potential by subtracting the potential of the anode from that of the cathode. This step is critical because it tells us which half-reaction should occur at the anode and which at the cathode under standard conditions, thus determining the flow of electrons within the cell and its overall voltage.

Reaction Quotient

The reaction quotient (\(Q\)) is a calculated value that indicates the relative amounts of products and reactants present during a chemical reaction at a given moment. For the electrochemical cell, \(Q\) is determined by the ratios of the concentrations of the reactants and products raised to the power of their stoichiometric coefficients.

In the given exercise, the reaction quotient is calculated by dividing the concentration of \(\mathrm{Cu}^+\) by the square of the concentration of \(\mathrm{I}^-\), reflecting the stoichiometry of the cell reactions. The reaction quotient is pivotal when using the Nernst equation to calculate the cell potential under non-standard conditions. A change in the reaction quotient reflects how the reaction shifts away from its standard state, therefore affecting the cell potential as well.