When hot and concentrated, sulfuric acid is a fairly strong oxidizing agent. Write a balanced net ionic equation for the oxidation of metallic copper to Cu(II) ion by hot concentrated H2SO4, in which the sulfur is reduced to SO2.
Cu ==> Cu^2+ + 2e
H2SO4 + 2H^+ + 2e ==> SO2 + 2H2O
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There are the two half equations. Add them for the total net ionic equation. If you wish to make a molecular equation, you will need to add some extra sulfate ions to form CuSO4.
To write a balanced net ionic equation for the oxidation of metallic copper (Cu) to Cu(II) ion by hot concentrated sulfuric acid (H2SO4), in which sulfur is reduced to sulfur dioxide (SO2), follow these steps:
Step 1: Write the unbalanced equation for the reaction.
Cu + H2SO4 → Cu(II) + SO2 + H2O
Step 2: Identify the balanced half-reactions for oxidation and reduction.
Oxidation Half-Reaction:
Cu → Cu(II) + 2e-
Reduction Half-Reaction:
H2SO4 + 2e- → SO2 + 2H+ + 2O2-
Step 3: Balance the number of electrons transferred in both half-reactions.
To balance the electrons in the oxidation half-reaction, multiply the reduction half-reaction by 2.
2(Cu → Cu(II) + 2e-) + H2SO4 → SO2 + 2H+ + 2O2-
Now, the number of electrons lost in the oxidation half-reaction matches the number of electrons gained in the reduction half-reaction.
Step 4: Cancel out any species that appear on both sides of the equation.
2(Cu) + 2(H2SO4) → (SO2) + 2(H+) + 2(O2-) + H2O
Step 5: Write the net ionic equation, which eliminates the spectator ions.
Net Ionic Equation:
Cu + H2SO4 → SO2 + H2O
This net ionic equation represents the oxidation of metallic copper to Cu(II) ion by hot concentrated H2SO4, in which sulfur is reduced to SO2.
To write a balanced net ionic equation for the oxidation of metallic copper to Cu(II) ion by hot concentrated H2SO4, we need to identify the oxidation and reduction half-reactions.
1. Oxidation half-reaction (loss of electrons):
Copper metal (Cu) is oxidized to Cu(II) ion (Cu^2+):
Cu(s) → Cu^2+(aq) + 2e^-
2. Reduction half-reaction (gain of electrons):
Sulfur in sulfuric acid (S in H2SO4) is reduced to sulfur dioxide (SO2):
S(s) → SO2(g)
3. Multiply the half-reactions by the appropriate coefficients so that the electron transfer is balanced. Since the number of electrons transferred in the oxidation half-reaction is 2, we need to multiply the reduction half-reaction by 2:
2S(s) → 2SO2(g)
4. The next step is to add the two equations together, canceling out any species that appear on both sides of the equation. In this case, the Cu(II) ion appears on both sides:
Cu(s) + 2S(s) → Cu^2+(aq) + 2SO2(g)
5. Finally, write the net ionic equation by removing the spectator ions, which are ions that do not participate in the reaction. In this case, there are no spectator ions, so the final net ionic equation is the same as the complete ionic equation:
Cu(s) + 2S(s) → Cu^2+(aq) + 2SO2(g)