Question
For each redox reaction shown below, identify the substance oxidized, the substance reduced, and, if any, spectator ions.
a. H₂ + F₂ → 2HF
b. 4Al(s) + 3O₂ → 2Al₂0₃ (s)
c. Mn0₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂0
d. Cl₂ (aq) + 2KI (aq) → I₂ (s) + 2KCl (aq)
a. H₂ + F₂ → 2HF
b. 4Al(s) + 3O₂ → 2Al₂0₃ (s)
c. Mn0₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂0
d. Cl₂ (aq) + 2KI (aq) → I₂ (s) + 2KCl (aq)
Answers
a. H₂ + F₂ → 2HF
Oxidation is the loss of electrons; reduction is the gain of electrons.
H on the left is 0 oxidation state; on the right H is +1 each. To get from zero to +1 it must have lost an electron so H is oxidized, then F mujst be reduced but you should confirm that. There are no spectator ion.
c. MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
Mn on the left is 4+ and on the right it is 2+ soit must have gained 2 electrons so Mn is reduced. Cl^- goes from 1- to 0 so that is a loss of electrons or oxidation. H^+, O^2- and part of the Cl^- are spectator ions. I'll leave the other two for you but I shall be happy to check your answers if you post them.
Oxidation is the loss of electrons; reduction is the gain of electrons.
H on the left is 0 oxidation state; on the right H is +1 each. To get from zero to +1 it must have lost an electron so H is oxidized, then F mujst be reduced but you should confirm that. There are no spectator ion.
c. MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O
Mn on the left is 4+ and on the right it is 2+ soit must have gained 2 electrons so Mn is reduced. Cl^- goes from 1- to 0 so that is a loss of electrons or oxidation. H^+, O^2- and part of the Cl^- are spectator ions. I'll leave the other two for you but I shall be happy to check your answers if you post them.
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