Question

Biofuel Cell At the cathode of a biofuel cell, Fe \(^{3+}\) in potassium hexacyanoiron(III) \(\left(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\right)\) is reduced to \(\mathrm{Fe}^{2+}\) in potassium hexacyanoiron (II) \(\left(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\right)\) At the anode, reduced nicotinamide-adenine-dinucleo-tide (NADH) is oxidized to NAD+. Use the following standard reduction potential to determine the potential of the cell. $$ \begin{array}{ll}{\mathrm{NAD}^{+}+\mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{NADH}} & {E^{0}=-0.320 \mathrm{V}} \\\ {\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+1 \mathrm{e}^{-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}} & {E^{0}=+0.36 \mathrm{V}}\end{array} $$

Short answer

The potential of the biofuel cell is 0.04V.

Step-by-step solution

Write the half-cell reactions correctly for the given biofuel cell

We have the following half-cell reactions: At the cathode (reduction): \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} + 1 \mathrm{e}^- \rightarrow \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\), with \(E^0 = +0.36V\) At the anode (oxidation): \(\mathrm{NADH} \rightarrow \mathrm{NAD}^{+} + \mathrm{H}^+ + 2 \mathrm{e}^{-}\), with \(E^0 = -0.320V\)

Reverse the oxidation half-cell reaction to match the cell notation and add both reactions

In a biofuel cell, NADH is actually oxidized to NAD+ at the anode. So, we need to reverse the given half-cell reaction for NADH: \(\mathrm{NAD}^{+} + \mathrm{H}^+ + 2 \mathrm{e}^- \cancel{\rightarrow}\cancel{\leftarrow} \mathrm{NADH}\), with \(E^0 = -(-0.320V) = +0.320V\) Now, we can add the two half-cell reactions: Cathode (reduction): \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} + 1 \mathrm{e}^- \rightarrow \left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) Anode (oxidation): \(\mathrm{NADH} \rightarrow \mathrm{NAD}^{+} + \mathrm{H}^+ + 2 \mathrm{e}^-\)

Balance half-cell reactions

We need to balance the electron transfer between the two half-cell reactions. To do this, we will multiply the first reaction by 2: Cathode (reduction): \(2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} + 2 \mathrm{e}^- \rightarrow 2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) Anode (oxidation): \(\mathrm{NADH} \rightarrow \mathrm{NAD}^{+} + \mathrm{H}^+ + 2 \mathrm{e}^-\) Now both reactions have two electrons being transferred. We can now add the two half-cell reactions together: Cell reaction: \(\mathrm{NADH} + 2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{NAD}^{+} + \mathrm{H}^+ + 2\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\)

Calculate the cell potential

Using the balanced half-cell reactions and their standard reduction potentials, we can calculate the cell potential using the Nernst equation: \(E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = (0.36V) - (0.320V) = 0.04V\) The potential of the biofuel cell is 0.04V.

Key concepts

Standard Reduction Potential

Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and be reduced. It is represented by the symbol \(E^0\) and measured in volts. A more positive \(E^0\) means a greater tendency to gain electrons and undergo reduction.

In biofuel cells, standard reduction potentials are essential to determine the voltage and effectiveness of the cell. For the reaction at the cathode in our example, the conversion of \([\mathrm{Fe}(\mathrm{CN})_6]^{3-}\) to \([\mathrm{Fe}(\mathrm{CN})_6]^{4-}\) has an \(E^0\) of \(+0.36V\). This indicates that the reaction strongly favors reduction.

Conversely, the reaction at the anode involves NADH being oxidized to NAD\(^+\), with a standard reduction potential of \(-0.320V\) when considered in reverse for reduction. When these values are used to evaluate the overall cell potential, they shed light on the feasibility and direction of the electrochemical reaction.

• Positive \(E^0\) value: Strong tendency to be reduced.
• Negative \(E^0\) value: Strong tendency to be oxidized as given.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, are chemical processes where electrons are transferred between substances. These reactions are vital for biofuel cells, as they facilitate the conversion of chemical energy into electrical energy.

In a typical biofuel cell, two half-reactions occur: one at the anode and one at the cathode. During oxidation at the anode, NADH is transformed into NAD\(^+\), donating electrons:
\[\mathrm{NADH} \rightarrow \mathrm{NAD}^{+} + \mathrm{H}^+ + 2 \mathrm{e}^-\]
At the cathode, reduction occurs where \([\mathrm{Fe}(\mathrm{CN})_6]^{3-}\) gains electrons:
\[2[\mathrm{Fe}(\mathrm{CN})_6]^{3-} + 2 \mathrm{e}^- \rightarrow 2[\mathrm{Fe}(\mathrm{CN})_6]^{4-}\]

Balancing electrons is crucial to ensure that the number of electrons lost equals the number gained, allowing the reaction to proceed smoothly.

• Anode: Site of oxidation where electrons are released.
• Cathode: Site of reduction where electrons are gained.

Nernst Equation

The Nernst equation allows us to calculate the cell potential under non-standard conditions. It accounts for the effect of concentration, pressure, and temperature on the cell potential. For the perfect functioning of a biofuel cell, understanding how to apply this equation is key.

The standard form of the Nernst equation is:
\[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\]
Where:

• \(E_{cell}\) is the cell potential at non-standard conditions.
• \(E^0_{cell}\) is the standard cell potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of moles of electrons transferred.
• \(F\) is Faraday’s constant.
• \(Q\) is the reaction quotient.

For our reaction, using the balanced half-cell potentials, the standard cell potential \(E^0_{cell}\) was calculated by taking the difference between the cathode and anode standard reduction potentials, resulting in \(0.04V\). The Nernst equation then helps adjust this value based on real-world conditions, optimizing the cell's efficiency.