What is the value of Eau for the given reaction with a standard reduction potential of 1.260 V and concentrations of 2A1 + 30d (q) + 3Cd (s) + 2A1" (24)?
The value of E for the given reaction is 1.260 V. To find the value of Eau, we need to consider the half-reaction of Au. The standard reduction potential for Au is 1.498 V, and we can calculate Eau using the Nernst equation.
Calculating E for Au Reduction
The balanced half-reaction for the reduction of Au ions is:
Au^3+ + 3e^- -> Au (s).
The standard reduction potential Ered for this half-reaction is 1.498 V.
Using the Nernst equation, E = Ered - (0.0592/n) log([Au^3+]/[Au]), where n is the number of electrons involved in the half-reaction, we can calculate the value of Eau by substituting the known values into the equation.
By plugging in the values:
E = 1.498 V - (0.0592/3) log(30/1) = 1.498 V - (0.01973) log(30)
E = 1.498 V - (0.01973) (1.47712) = 1.498 V - 0.02921 = 1.46879 V
Therefore, the value of Eau for the given reaction with the concentrations shown is approximately 1.46879 V.