Identify which substances are oxidized and which are reduced for the following reaction. Provide answers through steps by step
Zn + 2HCl = H2 + ZnCl
2
Step 1: Write the oxidation numbers for each element in the reactants and products:
Zn: 0
H: +1 (in HCl and H2)
Cl: -1 (in HCl and ZnCl2)
Step 2: Determine how the oxidation numbers change from reactants to products.
Zn goes from 0 to +2, a loss of electrons. This means Zn is oxidized.
H goes from +1 to 0, a gain of electrons. This means H is reduced.
Cl goes from -1 to 0, a gain of electrons. This means Cl is reduced.
Step 3: Write the half-reactions showing the oxidation and reduction separately:
Oxidation half-reaction:
Zn -> Zn2+ + 2e-
Reduction half-reaction:
2H+ + 2e- -> H2
Step 4: Check that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.
The oxidation half-reaction loses 2 electrons, while the reduction half-reaction gains 2 electrons. They balance each other out.
Therefore, Zn is oxidized and H and Cl are reduced in the reaction:
Zn + 2HCl -> H2 + ZnCl2
Step 1: Write down the half-reactions for oxidation and reduction.
In this reaction, zinc (Zn) reacts with hydrochloric acid (HCl) to produce hydrogen gas (H2) and zinc chloride (ZnCl).
The half-reaction for oxidation can be written as:
Zn → Zn2+ + 2e-
The half-reaction for reduction can be written as:
2H+ + 2e- → H2
Now, we can identify the substances that are being oxidized and reduced:
- Zinc (Zn) is oxidized because it loses electrons and forms Zn2+ ions.
- Hydrogen ions (H+) from hydrochloric acid are reduced because they gain electrons and form hydrogen gas (H2).