Question 13.EP.2: Suppose that you work for a company that designs the drive m...
Suppose that you work for a company that designs the drive mechanisms for large ships. The materials in this mechanism will obviously come into contact with environments that enhance corrosion. To estimate the difficulties that corrosion might cause, you decide to build a model electrochemical cell using electrolyte concentrations that might be present in your system when it is in service. Assume that you have a cell that has an iron(II) concentration of 0.015 M and an H^{+} concentration of 1.0 × 10^{−3} M. The cell temperature is 38°C, and the pressure of hydrogen gas is maintained at 0.04 atm. What would the cell potential be under these conditions?
Strategy
This problem defines nonstandard conditions that must be addressed using the Nernst equation. Virtually anytime you are given concentrations of electrolytes present in a cell (other than 1 M), you need this equation. This problem also presents the challenge of identifying the reactions involved. Iron will be the anode, but we will need to scan the table of standard reduction potentials to identify a possible cathode reaction. The most likely suspect is the reduction of H^{+} to H_{2}. Once we know both half-reactions, we can calculate the standard cell potential to fill in the appropriate values in the Nernst equation.
Anode reaction: Fe^{2+}(aq) + 2 \ e^{−} \longrightarrow Fe(s) E° = −0.44 V
Cathode reaction: 2 \ H^{+}(aq) + 2 \ e^{−} \longrightarrow H_{2}(g) E° = 0.00 V
These reactions tell us two things: first, the standard cell potential will be
E° = 0.00 V − (−0.44 V) = 0.44 V
Second, there are two electrons transferred in the overall redox reaction:
Fe(s)+2 \ H^{+}(aq)\longrightarrow H_{2}(g)+Fe^{2+}(aq)
These facts, plus the values given in the problem and those of the constants allow us to use the Nernst equation to find the cell potential:
Discussion
The correction made in the cell potential is rather small, even though we are far from standard conditions. But what happens as the reaction proceeds? The concentration of H^{+} decreases, and that of Fe^{2+} increases. So the potential is not a constant—ln Q becomes larger (a smaller number in the denominator, larger in the numerator), until the potential falls to zero (at least in principle). This rather complicated behavior is possible in a model that simplifies the real world situation, so we can see that it can be vital to account for nonstandard conditions for applications outside of the laboratory.