A student is studying the electrochemical behavior of a certain metal ion in a solution of varying pH values. They have measured the anodic and cathodic Tafel slopes for the electrode reaction at each pH. The anodic Tafel slope values range from 60 mV/decade to 90 mV/decade, and the corresponding cathodic Tafel slope values range from 50 mV/decade to 70 mV/decade. Is this metal ion undergoing a one-electron or a two-electron transfer reaction? Justify your answer.

Question

A student is studying the electrochemical behavior of a certain metal ion in a solution of varying pH values. They have measured the anodic and cathodic Tafel slopes for the electrode reaction at each pH. The anodic Tafel slope values range from 60 mV/deca

Answer 1

To determine whether the metal ion is undergoing a one-electron or a two-electron transfer reaction, we can use the Tafel equation and the relationship between the Tafel slopes and the electron transfer number  n . The Tafel equation is given by: = b * log10 j / j0 where  is the overpotential, b is the Tafel slope, j is the current density, and j0 is the exchange current density.The Tafel slope  b  is related to the electron transfer number  n  by the following equation:b =  2.303 * R * T  /  n * F where R is the gas constant  8.314 J/molK , T is the temperature in Kelvin, n is the electron transfer number, and F is the Faraday constant  96,485 C/mol .For a one-electron transfer reaction, n = 1, and for a two-electron transfer reaction, n = 2. We can use the given Tafel slope values to estimate the electron transfer number.Let's consider the average anodic Tafel slope value, which is 75 mV/decade  midpoint between 60 and 90 mV/decade . Assuming a temperature of 298 K  25C , we can calculate the electron transfer number  n  as follows:75 mV/decade =  2.303 * 8.314 J/molK * 298 K  /  n * 96,485 C/mol 75 * 10^-3 V/decade =  2.303 * 8.314 * 298  /  n * 96,485 n  1.06Now, let's consider the average cathodic Tafel slope value, which is 60 mV/decade  midpoint between 50 and 70 mV/decade :60 mV/decade =  2.303 * 8.314 J/molK * 298 K  /  n * 96,485 C/mol 60 * 10^-3 V/decade =  2.303 * 8.314 * 298  /  n * 96,485 n  1.29Both calculations give n values close to 1, which indicates that the metal ion is undergoing a one-electron transfer reaction. The slight deviation from n = 1 could be due to experimental errors or other factors affecting the Tafel slope values. However, the overall trend suggests a one-electron transfer process.