If the overpotential of an electrolysis process is increased from 0.5 V to 0.6 V, then the ratio of current densities (In \(\frac{\int0.6 }{\int0.5}\)) of the electrolysis will be equal to (given transfer co - efficient = 0.5)
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.
Important Questions from Electrochemistry
- At $298 \, K$, given the standard electrode potentials: $E^\circ_{Cu^{2+}/Cu} = 0.34 \, V$, $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$, $E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$, and $E^\circ_{Ag^{+}/Ag} = 0.80 \, V$.
Based on these values, which of the following reactions is NOT expected to occur spontaneously under standard conditions? You are given three metals 'X', 'Y' and 'Z'. Metal 'X' is found to react with an aqueous solution of both YSO 4and ZSO 4whereas metal 'Z' is found to react only with aqueous solution of YSO 4. Based on these observations, select the correct statement from the following.
The chemical potential (μ) of a 2 molar Na2SO4 solution is expressed in terms of mean ionic activity co - efficient (γ±) as
Which statement is true for electrochemical series?
The E°(M+/M) of the cell, SHE||MX|M can be obtained from the plot of (E cell is the cell potential and m is the molality of ideal dilute solution of MX)
