Question

Calculate E_cell for the following electrochemical cell:

Pt/Fe²⁺ (0.0725M), Fe³⁺ (0.110M)// MnO4^- (0.0822M), H⁺ (1.00M), Mn²⁺ (0.0205M) /Pt

Answer 1

Solution :

From above cell we write oxidation and reduction half and balance.

MnO4^- (aq) --- > Mn²⁺ (aq)

Balancing :

MnO4^- (aq) + 8 H⁺(aq) --- > Mn²⁺ (aq) + 4H2O (l)

Charge :

MnO4^- (aq) + 8 H⁺(aq)+5 e- --- > Mn²⁺ (aq) + 4H2O (l)

Oxidation half :

Fe2+ ----- > Fe3+

Fe2+ ----- > Fe3+ +1e-

Lets make electrons equal

5Fe2+ ----- > 5Fe3+ +5e-

Add these two half

MnO4^- (aq) + 8 H⁺(aq)+5 e- --- > Mn²⁺ (aq) + 4H2O (l)

5Fe2+ ----- > 5Fe3+ +5e-

--------------------------------------------------------------------

MnO4^- (aq) + 5Fe²⁺(aq) +8H⁺(aq) --------- > 5Fe³⁺(aq) + Mn²⁺ (aq) + 4 H2O (l)

This is overall reaction in this given cell.

Oxidation half is at anode and reduction half is at cathode

Lets calculate E⁰ cell = E cathode – E anode

MnO4^- (aq) + 8 H⁺(aq)+5 e- --- > Mn²⁺ (aq) + 4H2O (l)   E⁰ =1.51 V

5Fe2+ ----- > 5Fe3+ +5e-    E⁰cell = 0.77 V

E⁰cell = 1.51 – 0.77 = 0.74 V

Now we use Nernst equation to calculate E cell

E_cell = E⁰_cell –    ( 0.0592 V / n ) * logQ

Here Q is reaction quotient.

Number of electrons transferred = 5

Lets use above given concentrations and n to find Ecell

E_cell =0.74 V –( 0.0592 V / 5 ) log([Fe³⁺]⁵ [Mn²⁺ ] /[MnO4^-][Fe²⁺]⁵[H⁺]⁸)

E_cell =0.74 V –( 0.0592 V / 5 ) log([0.110]⁵ [0.0205] /[0.0822][0.0725]⁵[1.0]⁸)

Ecell = 0.736 V