Question

Calculate the Ecell value at 298K for the cell based on the reaction

Fe³⁺(aq)+ Cr²⁺(aq) --> Fe²⁺(aq)+Cr³⁺(aq)

when [Fe³⁺] = [Cr²⁺] = 1.50x10^-3 M and [Fe²⁺]= [Cr³⁺] = 2.5x10^-4 M?

Answer 1

Solution :-

Fe³⁺(aq)+ Cr²⁺(aq) --> Fe²⁺(aq)+Cr³⁺(aq)

For the given reaction Fe^3+ is reduced to Fe^2+ (cathode reaction) and Cr^2+ is oxidized to Cr^3+ (anode reaction )

Now lets calculate the standard cell potential using the standard reduction potential values

Cr^3+ + 1e-   ---- > Cr^2+            Eo = -0.41 V

Fe^3+ + 1e- ----- > Fe^2+          Eo = 0.77 V

Now lets calculate the Eo cell using the standard reduction potentials

Eo cell = Eo cathode – Eo anode

Eo cell = 0.77 V – (-0.41 V)

             = 1.18 V

Now lets calculate the cell potentials at the given concentrations using the Eo cell

when [Fe³⁺] = [Cr²⁺] = 1.50x10^-3 M and [Fe²⁺]= [Cr³⁺] = 2.5x10^-4 M?

Nernst equation

E cell = Eo cell – (0.0592/n) log Q

Q= [product] / [reactant]

n= number of electrons transferred = 1

Now lets put the values in the formula

E cell = 1.18 V – (0.0592 /1) *log [(2.5x10^-4)( 2.5x10^-4)]/[( 1.50x10^-3)( 1.50x10^-3)]

E cell = 1.18 V – (-0.26 V)

E cell = 1.44 V

Therefore Cell potential at the given concentrations is 1.44 V