Electrolysis Overpotential and Current Density

The relationship between overpotential and current density in an electrochemical process is described by the Butler-Volmer equation. For large overpotentials, this equation simplifies to the Tafel equation.

The anodic Tafel equation can be written in terms of current density \(i\) as:

\( i = i_0 \exp\left(\frac{\alpha n F \eta}{RT}\right) \)

Where:

• \(i\) is the current density
• \(i_0\) is the exchange current density
• \(\alpha\) is the transfer coefficient
• \(n\) is the number of electrons transferred in the rate-determining step
• \(F\) is the Faraday constant
• \(R\) is the ideal gas constant
• \(T\) is the absolute temperature
• \(\eta\) is the overpotential

We are given the transfer coefficient \(\alpha = 0.5\). The overpotential changes from \(\eta_1 = 0.5\) V to \(\eta_2 = 0.6\) V. We need to find the ratio \(\ln \left(\frac{i_{0.6}}{i_{0.5}}\right)\), where \(i_{0.5}\) and \(i_{0.6}\) are the current densities at \(\eta = 0.5\) V and \(\eta = 0.6\) V respectively.

Using the Tafel equation, the current density at \(\eta_1 = 0.5\) V is:

\( i_{0.5} = i_0 \exp\left(\frac{\alpha n F \times 0.5}{RT}\right) \)

The current density at \(\eta_2 = 0.6\) V is:

\( i_{0.6} = i_0 \exp\left(\frac{\alpha n F \times 0.6}{RT}\right) \)

Now, let's find the ratio of the current densities:

\( \frac{i_{0.6}}{i_{0.5}} = \frac{i_0 \exp\left(\frac{\alpha n F \times 0.6}{RT}\right)}{i_0 \exp\left(\frac{\alpha n F \times 0.5}{RT}\right)} \)

\( \frac{i_{0.6}}{i_{0.5}} = \exp\left(\frac{\alpha n F \times 0.6}{RT} - \frac{\alpha n F \times 0.5}{RT}\right) \)

\( \frac{i_{0.6}}{i_{0.5}} = \exp\left(\frac{\alpha n F (0.6 - 0.5)}{RT}\right) \)

\( \frac{i_{0.6}}{i_{0.5}} = \exp\left(\frac{\alpha n F \times 0.1}{RT}\right) \)

Next, we take the natural logarithm of the ratio:

\( \ln\left(\frac{i_{0.6}}{i_{0.5}}\right) = \ln\left[\exp\left(\frac{\alpha n F \times 0.1}{RT}\right)\right] \)

\( \ln\left(\frac{i_{0.6}}{i_{0.5}}\right) = \frac{\alpha n F \times 0.1}{RT} \)

We are given \(\alpha = 0.5\). The number of electrons transferred, \(n\), is not explicitly given, but the form of the options suggests that \(n=1\) is assumed for this calculation, as is common in simplified kinetics problems unless specified otherwise.

Substituting the values:

\( \ln\left(\frac{i_{0.6}}{i_{0.5}}\right) = \frac{0.5 \times 1 \times 0.1 F}{RT} \)

\( \ln\left(\frac{i_{0.6}}{i_{0.5}}\right) = \frac{0.05 F}{RT} \)

This result matches one of the given options.