Balance the redox reaction below using the half-reaction method. Fe3+(aq) + Gd(s) Fe2+(aq) + Gd3+(aq)

Question

Balance the redox reaction below using the half-reaction method.

1. Fe³⁺(aq) + Gd(s)  Fe²⁺(aq) + Gd³⁺(aq)

(a) To show your method, write the balanced half reactions below. Use the smallest integer coefficients possible and show electrons as e^-. If a box is not needed, leave it blank. (Coefficients of 1 are not needed).

Oxidation half-reaction:

+

+

Reduction half-reaction:

+

+

(b) To show your balanced equation, enter an integer in each of the boxes but if the integer is "1", you may leave the box blank. Use the smallest integer coefficients possible.

Fe3+(aq) +   Gd(s)

Fe2+(aq) +  Gd3+(aq)

Answer 1

a. RedOx reaction:

Reduction half + Oxidation half = RedOx full

OIL: Oxidation Is Loss of electron(s)

               Gd(s) – 3 e -->  Gd³⁺(aq)                             loss of 3 electrons

RIG: Reduction Is Gain of electron(s)

               Fe³⁺(aq) + e --->  Fe²⁺(aq)                              gain of 1 electron

b.      To write a balanced RedOx equation, number of electrons lost and gained must be equal; in this case, Reduction Half Reaction has to be re-written as:

3 Fe³⁺(aq) + 3 e --->  3 Fe²⁺(aq)  gain of 3 electrons

Then put the Oxidation and Reduction together, and cancel electrons,

3 Fe³⁺(aq) + Gd(s)  ---> 3 Fe²⁺(aq) + Gd³⁺(aq)