Calculate the pH of the cathode compartment for the following reaction given Ecell = 3.01 V when [Cr3+] = 0.15 M, [Al3+] = 0.30 M, and [Cr2O72-] = 0.55 M.
2 Al(s) + Cr2O72-(aq) + 14 H+(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
-0.24925=log {h+)
thanks
To calculate the pH of the cathode compartment for the given reaction, we need to use the Nernst equation. The Nernst equation relates the standard cell potential (E°cell) to the actual cell potential (Ecell) as follows:
Ecell = E°cell - (RT/nF) * ln(Q)
In this equation:
- Ecell is the actual cell potential
- E°cell is the standard cell potential
- R is the ideal gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced redox equation
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient
For the given reaction:
2 Al(s) + Cr2O72-(aq) + 14 H+(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
The number of electrons transferred (n) is 6, as seen from the balanced equation.
To calculate Q, we need to find the concentrations of the species involved. Given concentrations are:
[Cr3+] = 0.15 M
[Al3+] = 0.30 M
[Cr2O72-] = 0.55 M
For Q, we use the concentrations of the products (Al3+ and Cr3+) raised to their respective stoichiometric coefficients and divide by the concentrations of the reactants (Cr3+ and Al3+) raised to their respective stoichiometric coefficients:
Q = ([Al3+]^2 * [Cr3+]^2) / ([Cr2O72-] * [H+]^14)
Now we can substitute the values of Q, E°cell, R, T, n, and F into the Nernst equation and solve for pH.
Keep in mind that pH is defined as the negative logarithm of the concentration of H+ ions:
pH = -log[H+]
So let's put it all together and calculate the pH of the cathode compartment.