The balanced half-reaction for the product that forms at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻ → I₂ + 2e- and 2H₂O → O₂ + 4H⁺ + 4e⁻
<h3>What is Balanced Chemical Equation ?</h3>
The equation during which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation is called balanced chemical equation.
Now write the equation for FeI₂
At cathode:
Fe⁺² + 2e⁻ → Fe(s) Eo = - 0.44 V
2H₂O + 2e⁻ → H₂ + 2OH⁻ Eo = - 0.827 V
It is easy to decrease Fe⁺² ions than the water, the product which is formed at cathode is Iron.
At anode:
2I⁻ → I₂ + 2e- Eo = - 0.54 V
2H₂O → O₂ + 4H⁺ + 4e⁻ Eo = -1.23 V
O₂ gas formed at anode.
Thus from the above conclusion we can say that The balanced half-reaction for the product at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻ → I₂ + 2e- and 2H₂O → O₂ + 4H⁺ + 4e⁻.
Learn more about the Balanced chemical equation here: brainly.com/question/26694427
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