Question

The standard half-cell potential for the reaction
O2(g)+4H+(aq)+4e−→2H2O(l) is +1.229 V at 298.15 K.
The aO2 = 1.00 assuming that the aH+ is equal to the molality.

Part A

Calculate E for a 0.100-molal solution of H2SO4 for aO2 = 1.00 assuming that the aH+ is equal to the molality.

Part B

Calculate E for a 0.100-molal solution of H2SO4 for aO2= 1.00 using the measured mean ionic activity coefficient for this concentration from the data tables in the textbook.

Part C

How large is the relative error if the concentrations, rather than the activities, are used?

Answer 1

To find the electrochemical potential of a cell we use the Nerst equation which is given as:

Q=reaction quotient or simply it can be replaced with K the equilibrium constant.

n= number of electrons

F=Faradays Constant=96500C

From the given data we have the following reaction:

The standard electrode potential E0 is given as 1.229V.

To calculate the electrochemical potential of cell using its mean ionic activity we first need to calculate the mean ionic activity of the cell

Mean ionic activity =molality*mean activity coefficient(γ)

Here we are not given with the value of mean activity coefficicent (γ) so we are taking the standard value to be 0.80.

Value of electrochemical potential when molality of H⁺ is approximate to its activity[E]_molality=1.188

Value of electrochemical potential with mean ionic activity measurment for H⁺ [E]_activity=1.182

The percentage error would be very less as the first two digits of decimal are almost same.