Question

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}_{,} P_{\mathrm{H}_{2}}=\) \(0.95 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

Short answer

The emf of the voltaic cell under standard conditions is +0.771 V. Under the given non-standard conditions with \([Fe^{3+}] = 3.50\, M\), \(P_{H_2} = 0.95\, atm\), \([Fe^{2+}] = 0.0010\, M\), and pH in both half-cells is \(4.00\), the emf of the cell is approximately +0.971 V.

Step-by-step solution

Part (a): Find the emf of the cell under standard conditions

First, we need to identify the reduction and oxidation half-reactions. Oxidation is the loss of electrons, and reduction is the gain of electrons. Reduction: \(2Fe^{3+}(aq) + 2e^- → 2Fe^{2+}(aq)\) (Fe being reduced from +3 to +2) Oxidation: \(H_2(g) → 2H^+(aq) + 2e^-\) (H_2 being oxidized losing electrons) Next, we will look up the standard electrode potentials for these half-reactions: Reduction: \(Fe^{3+} + e^- → Fe^{2+}\) with \(E^0_{Red}=+0.771\, V\) Oxidation: \(H_2 → 2H^+ + 2e^-\) with \(E^0_{Ox}=0\, V\) (standard hydrogen electrode) Now, we can calculate the standard emf, \(E_{cell}^0\): \(E_{cell}^0 = E_{Red}^0 - E_{Ox}^0 = +0.771\, V - 0\, V= +0.771\, V\) So the emf of this cell under standard conditions is +0.771 V.

Part (b): Find the emf of the cell under non-standard conditions

Now we will find the emf of the cell under the given concentrations and pressures. For this, we will use the Nernst Equation: \(E_{cell}=E^0_{cell} - \frac{RT}{nF} \ln Q\) where E is the emf under non-standard conditions, \(E^0_{cell}\) is the standard emf calculated above, R is the universal gas constant (8.314 J/(mol·K)), T is the temperature (298 K under standard conditions), n is the number of electrons transferred (2 in this case), F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. First, let's find the reaction quotient, Q. In this case, Q can be written as: \(Q = \frac{[Fe^{2+}]^2[H^+]^2}{[Fe^{3+}]^2}\) Using the given concentrations and pressures, and knowing the pH, we can find the concentrations of each ion: \([Fe^{3+}] = 3.50\, M\) \(P_{H_2} = 0.95\, atm\) \([Fe^{2+}] = 0.0010\, M\) \(pH = 4.00 \Rightarrow [H^+] = 10^{-4}\, M\) Now we can plug these values into Q: \(Q = \frac{(0.0010)^2(10^{-4})^2}{(3.50)^2} = 8.16 \times 10^{-15}\) Next, we will use this value of Q in the Nernst Equation to find the non-standard emf: \(E_{cell} = +0.771 - \frac{8.314 \times 298}{2 \times 96485} \ln(8.16 \times 10^{-15})\) \(E_{cell} \approx +0.971\, V\) So the emf of this cell under the given non-standard conditions is approximately +0.971 V.

Key concepts

Standard Electrode Potentials

Understanding standard electrode potentials is crucial for predicting the direction of an electrochemical reaction and calculating the electromotive force (EMF) in a voltaic cell. Electrons move from regions of higher potential to lower potential in an electrochemical cell. The standard electrode potential, usually denoted by the symbol \(E^0\), is a measure of the intrinsic tendency of a species to lose or gain electrons, and it is referenced against the standard hydrogen electrode which is arbitrarily given a potential of \(0.00\text{ V}\).

These potentials are measured under standard conditions, which include a concentration of \(1 \text{ M}\) for aqueous solutions, a pressure of \(1 \text{ atm}\) for gases, and a temperature of \(25^{\text{o}}C\) or \(298 \text{ K}\). In the given exercise, the standard potential of \(Fe^{3+}/Fe^{2+}\) couple is given as \(+0.771\text{ V}\), indicating that iron(III) ions have a strong tendency to gain electrons and be reduced to iron(II) ions under standard conditions.

Nernst Equation

While the standard electrode potentials give us a good understanding of the cell voltage under ideal conditions, real-world scenarios require adjustments for non-standard conditions. This is where the Nernst equation becomes highly beneficial. The Nernst equation relates the standard cell potential, \(E^0_{cell}\), to the actual cell potential, \(E_{cell}\), under any given set of conditions. The equation considers the effect of temperature (T), the reaction quotient (Q), the number of electrons transferred in the reaction (n), the Faraday constant (F), and the universal gas constant (R).

The full Nernst equation for a reaction at temperature T is given by:\[\begin{equation}E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\text{,}\end{equation}\]In the exercise, we apply this equation to calculate the EMF of the cell at specific concentrations and pressures, which are different from standard conditions. This showcases how electrochemical reactions can be influenced by the environment and the concentrations of the involved species.

Reaction Quotient

The reaction quotient, \(Q\), plays an integral role in the Nernst equation as it accounts for the real-time concentrations or partial pressures of reactants and products in an electrochemical reaction. Essentially, it's akin to the equilibrium constant \(K\), but for a system that hasn't necessarily reached equilibrium. For the given reaction with the formula:\[\begin{equation}Q = \frac{[products]}{[reactants]}\text{,}\end{equation}\]the concentrations of the products are represented in the numerator, and the reactants occupy the denominator. In the context of the exercise, we calculate \(Q\) by considering the concentrations of \(Fe^{2+}\), \(H^+\), and \(Fe^{3+}\), which are affected by the cell's environment such as the pH level. By plugging the calculated \(Q\) value into the Nernst equation, we determine the EMF of the cell under non-standard conditions, accurately reflecting the actual behavior of the system.

Half-Reactions in Electrochemistry

Electrochemical cells operate on the principle of redox reactions, where oxidation and reduction processes occur simultaneously. To analyze such reactions, we break them down into half-reactions, each representing the oxidation or reduction event. In the current problem, the half-reactions are:\[\begin{equation}\text{Reduction (gain of electrons): } 2Fe^{3+}(aq) + 2e^- \rightarrow 2Fe^{2+}(aq)\end{equation}\]\[\begin{equation}\text{Oxidation (loss of electrons): } H_2(g) \rightarrow 2H^+(aq) + 2e^-\end{equation}\]Each half-reaction has an associated electrode potential. It's important that students thoroughly understand how to identify and write down half-reactions, because they represent the changes happening at each electrode. When connected, these half-reactions result in an overall net reaction in the electrochemical cell. By analyzing half-reactions, students can gain insights into the process of electron transfer and the chemistry underpinning the cell's generation of EMF.