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

(a) The emf of the cell under standard conditions is \(+0.77V\). (b) 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 \(4.00\), is approximately \(0.29V\).

Step-by-step solution

Write the balanced half-reactions and determine standard cell potential

We'll need to write the balanced half-reactions, find their standard potentials, and add them together to find the standard cell potential. Oxidation half-reaction (from \(Fe^{2+}\) to \(Fe^{3+}\)): \[Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^{-}\] Reduction half-reaction (from \(H^{+}(aq)\) to \(H_{2}(g)\)): \[2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)\] Look up the standard electrode potentials for these reactions in a table (assume 25ºC). The standard reduction potential for \(Fe^{3+} + e^{-} \rightarrow Fe^{2+}\) is \(+0.77V\). However, since our oxidation half-reaction is the reverse, we need to change the sign: \(E^0_{oxidation}=-0.77V\). The standard reduction potential for the reduction of hydrogen ions to form hydrogen gas is \(0V\), since the hydrogen electrode serves as a reference. So, our standard cell potential is: \[E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0V - (-0.77V) = +0.77V\]

Answer question (a): Standard EMF

Since the emf of the cell under standard conditions (concentrations of 1M and pressure of 1 atm) is equal to the standard potential, we have: emf under standard conditions = \(E^0_{cell} = +0.77V\)

Compute non-standard EMF using Nernst Equation

Now we will use the Nernst equation to find the non-standard EMF for the given concentrations and pressure. The Nernst equation is: \[E = E^0 - \frac{RT}{nF} \ln(Q)\] Where: - \(E\) is the non-standard EMF - \(E^0\) is the standard cell potential - \(R\) is the ideal gas constant, 8.314 J/mol˙K - \(T\) is the temperature in Kelvin, 298 K - \(n\) is the number of electrons transferred, 2 - \(F\) is the Faraday's constant, 96485 C/mol - \(Q\) is the reaction quotient Here, the reaction quotient Q is given by: \[Q = \frac{([Fe^{3+}]^2[P_{H_2}[H^{+}]^2)}{[H^{+}]^{2}\times[Fe^{2+}]}\] where the given concentrations, pH, and pressure: \( \begin{cases} & [Fe^{3+}] = 3.50M \\ & [Fe^{2+}] = 0.0010M \\ & \textrm{pH} = 4 \Rightarrow [H^+] = 10^{-4}M \\ & P_{H_2} = 0.95 \textrm{atm} \end{cases} \) We can simplify Q as: \[Q = \frac{[Fe^{3+}]^2}{[Fe^{2+}]}= \frac{(3.50M)^2}{0.0010M}\] Now, plug the values into the Nernst equation and solve for E: \[E = 0.77V - \frac{8.314 J/mol\cdot K \times 298 K}{2 \times 96485 C/mol} \ln \left(\frac{(3.50)^2}{0.0010}\right)\] \[E \approx 0.29V\]

Answer question (b): Non-standard EMF

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, is approximately \(0.29V\).

Key concepts

Standard Cell Potential

Understanding the standard cell potential is crucial when studying electrochemical cells. It is the measure of the potential difference between two half-cells in an electrochemical cell when all species are under standard conditions, namely a concentration of 1 M for aqueous species, a pressure of 1 atm for gases, and a common temperature of 25°C (298K).

For the reaction provided in the exercise, the standard cell potential can be determined by looking at the standard reduction potentials of the individual half-reactions involved and then combining them. The standard reduction potential for the half-reaction involving the reduction of iron (III) to iron (II) ions is +0.77 V, while the standard potential for the hydrogen electrode is 0 V. By reversing the iron half-reaction from reduction to oxidation, we change the sign of its potential. The standard cell potential (E^0_{cell}) is then, the difference between the potentials of the cathode and the anode:\[E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0V - (-0.77V) = +0.77V\].

Key takeaway: Standard cell potential enables us to predict the direction in which an electrochemical reaction will occur under standard conditions, as a positive value suggests a spontaneous reaction.

Nernst Equation

The Nernst equation is a mathematical representation that explains the effect of concentration, pressure, and temperature on the electrochemical cell potential. When the conditions deviate from the standard state, the Nernst equation is used to calculate the non-standard cell potential (E).

Given by the following formula:\[E = E^0 - \frac{RT}{nF} \ln(Q)\],where E^0corresponds to the standard cell potential, R is the gas constant, Tgoing on the temperature in Kelvin, nthe number of moles of electrons exchanged during the electrochemical reaction, and F the Faraday constant, the Nernst equation incorporates the reaction quotient, Q, to reflect the real-time composition of the reactants and products.

Applying the Nernst equation to our exercise, we can accurately determine the cell potential under non-standard conditions, incorporating the concentration of iron ions, hydrogen pressure, and the pH level.

Reaction Quotient

The reaction quotient (Q) is an expression that provides a snapshot of a reaction's composition at any moment in time. It has the same form as the equilibrium constant (K) but does not imply that the system is at equilibrium. Instead, Qtells you how far the system is from reaching equilibrium, by comparing the concentration of products to reactants.

Qis calculated by dividing the product of the concentrations of the products, each raised to the power of its coefficient in the balanced equation, by the product of the concentrations of the reactants, also raised to their respective coefficients. For gases, partial pressures are used instead of concentrations, and activities are used for solutions. Just like in the provided solution, knowing the reaction quotient is essential for applying the Nernst equation to calculate non-standard potentials or to predict the direction of the reaction.

Half-Reactions

In electrochemistry, we divide the overall redox reaction into two separate half-reactions: oxidation and reduction processes. This makes it easier to understand and analyze the chemistry happening within an electrochemical cell. An oxidation half-reaction involves the loss of electrons, whereas a reduction half-reaction is where the gain of electrons occurs.

In the exercise, we identify the iron reaction as the oxidation half-reaction, converting Fe^{2+}to Fe^{3+} by losing an electron. Conversely, the hydrogen reaction is the reduction half-reaction, where hydrogen ions gain electrons to form hydrogen gas. These half-reactions are crucial because they help us to establish the standard cell potential and make calculations using the Nernst equation more manageable. They also underpin the fundamental operations of all electrochemical cells, including batteries, fuel cells, and electrolysis setups.