Problem 46
Question
Biofuel Cell At the cathode of a biofuel cell, \(\mathrm{Fe}^{3+}\) 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-dinucleotide (NADH) is oxidized to NAD \(^{+} .\) Use the following standard reduction potential to determine the potential of the cell. $$ \begin{array}{ll}{\mathrm{NAD}^{+}+\mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{NADH}} & {E^{0}=-0.320 \mathrm{V}} \\\ {\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+1 \mathrm{e}^{-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}} & {E^{0}=+0.36 \mathrm{V}}\end{array} $$
Step-by-Step Solution
VerifiedKey Concepts
Half-Reactions
To identify the half-reactions:
- The reduction half-reaction at the cathode is: \[\mathrm{Fe}^{3+} + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+}\]
- The oxidation half-reaction at the anode is the reverse of the given reduction reaction, and it is:\[\mathrm{NADH} \rightarrow \mathrm{NAD}^+ + \mathrm{H}^+ + 2\mathrm{e}^-\]
Standard Reduction Potential
In our specific biofuel cell case, the given standard potentials are:
- For the cathode: \[E^0 = +0.36 \, \text{V}\]This value shows the ability of \[\left[\mathrm{Fe} (\mathrm{CN})_{6}\right]^{3-}\]to be reduced to \[\left[\mathrm{Fe} (\mathrm{CN})_{6}\right]^{4-}.\]
- For the anode: The reverse reaction has \[E^0 = -0.320 \, \text{V}\].Thus, \[\mathrm{NAD}^{+}\]can gain electrons to form \[\mathrm{NADH}\].But for oxidation, the sign changes to \[+0.320 \, \text{V}\].
Cell Potential Calculation
For the biofuel cell:
- The cathode reduction potential is \(+0.36 \, \text{V}\).
- After reversing the anode reaction, the oxidation potential becomes \(+0.320 \, \text{V}\).
NADH Oxidation
The oxidation reaction is:\[\mathrm{NADH} \rightarrow \mathrm{NAD}^+ + \mathrm{H}^+ + 2\mathrm{e}^-\]To determine how favorable this oxidation is, we rely on the standard reduction potential, which for the reverse reduction reaction is \(-0.320 \, \text{V}\). Flipping this for oxidation yields \(+0.320 \, \text{V}\). By losing electrons, NADH contributes to the overall potential generated by the cell and supports energy production.
Fe(III) Reduction
The half-reaction for this reduction is:\[\mathrm{Fe}^{3+} + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+}\]The standard reduction potential involved here is \(+0.36 \, \text{V}\), a positive value indicating that the reduction of Fe(III) to Fe(II) is favorable.
The presence of the hexacyanoferrate complexes \(\mathrm{K}_3\left[\mathrm{Fe} (\mathrm{CN})_6\right]\)and \(\mathrm{K}_4\left[\mathrm{Fe} (\mathrm{CN})_6\right]\)gives stability to the Fe ions during the reaction, enabling efficient electron transfer and ultimately contributing to the cell potential of the biofuel cell.