A balanced half reaction for the product that forms at anode is 2H₂O → O₂ + 4H⁺ + 4e⁻ and product that forms at cathode is Sn⁺² + 2e⁻ → Sn and 2H₂O + 2e⁻ → H₂ + 2OH⁻
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now write the half equation for
At cathode:
Sn⁺² + 2e⁻ → Sn
2H₂O + 2e⁻ → H₂ + 2OH⁻
Oxidizing agent are Sn⁺² and H₂O
At anode:
2H₂O → O₂ + 4H⁺ + 4e⁻
Anion SO₄⁻² cannot be oxidized further because it is already in its highest oxidation state which is +6.
Thus from the above conclusion we can say that A balanced half reaction for the product that forms at anode is 2H₂O → O₂ + 4H⁺ + 4e⁻ and product that forms at cathode is Sn⁺² + 2e⁻ → Sn and 2H₂O + 2e⁻ → H₂ + 2OH⁻
Learn more about the Balanced chemical equation here: brainly.com/question/26694427
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