In the electrochemical cell using the redox reaction below, the oxidation half-reaction is sn4+ (aq) + fe (s) → sn2+ (aq) + fe2+ (aq)

Fe(s)   ⇒ Fe²⁺  + 2e⁻ Oxidation

Explanation:

Oxidation is the loss of electrons, so what we have to do here is to determine the change in oxidation of the atoms in the reaction and identify the species which loses electrons.

Sn⁴⁺ + 2e⁻  ⇒  Sn²⁺   Reduction

Fe(s)   ⇒ Fe²⁺  + 2e⁻ Oxidation

Then the oxidation half-reaction is:

Fe(s)   ⇒ Fe²⁺  + 2e⁻ Oxidation

dynamic equilibrium is the answer.