Question

Consider the following UNBALANCED reaction IN ACIDIC SOLUTION:

Fe²⁺(aq) + MnO₄^-(aq) ⟶ Fe³⁺(aq) + Mn²⁺(aq)

a. (5) Species that is oxidized (be specific – i.e. identify which atom and if there are multiple atoms with different charges, identify the correct one)

b. (5) Species that is reduced (same instructions as above)

c. (10) Full Balanced oxidation ½ reaction (‘full’ means with regard to mass (atoms) and charges.)

d. (10) Full Balanced reduction ½ reaction

e. (9) Complete Balanced Reaction

Using your balanced reaction above, provide the following information.

f. (3) Cell potential for the oxidation half reaction (E^o_ox)

g. (2) Cell potential for the reduction half reaction (E^o_red)

h. (7) E^o_cell

i. (7) Δ G o

j. (7) K

Answer 1

OXIDATION HALF Fe2+ = Fe3+(+2 to +3) Balance e: Fe2+ = Fe3++ e- 5Fe2+ = 5Fe3++ 5e-

REDUCTION HALF MnO4- = Mn2+(+7 to +2) Balance O:MnO4-= Mn2++ 4H2O Balance H: MnO4-+ 8H+=Mn2++4H2O Balance e: MnO4-+8H++5e-= Mn2++4H2O MnO4-+8H++5e-= Mn2++4H2O

5Fe²⁺+MnO₄^- + 8H⁺ = 5Fe³⁺+Mn²⁺ + 4H₂O

E⁰_anode = -0.77 V( 5Fe²⁺ = 5Fe³⁺+ 5e^-)

E⁰_cathode = 1.51 V( MnO₄^-+8H⁺+5e-= Mn²⁺+4H₂O)

E⁰_cell = 1.51 - -0.77 = 2.28 V

ΔG= -nFE⁰_cell = - 5x96500x2.28 = -1100100 J = -1100.1 KJ