Answer:
a)
Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)
#p⁺ 30 2 30 2
#e⁻ 30 0 28 2
b) 32
c) 30
d) Zn
Step-by-step explanation:
There is some info missing. I will add the complete question.
Consider the following oxidation/reduction reaction.
Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)
#p⁺ 2
#e⁻ 2
a) Fill in the number of protons and electrons for each product and reactant (two boxes have been filled in for you).
b) Verify that the number of protons on the left side of the chemical equation is equal to the number of protons of the right side. Show your work.
c) Verify that the number of electrons on the left side of the chemical equation is equal to the number of electrons of the right side. Show your work.
d) Which substance is being oxidized?
a) The atomic number of Zn is 30 so it will have 30 protons. Since it ts neutral, it will also have 30 electrons. Zn²⁺ will also have 30 protons but it lost 2 electrons, so it has 28 electrons. The atomic number of H⁺ is 1, so each H atom will have 1 proton (2 in total). But since H has 1 electron, and H⁺ lost 1 electron, H⁺ will have 0 electrons. The complete chart is:
Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)
#p⁺ 30 2 30 2
#e⁻ 30 0 28 2
b) The total number of protons on the left side is: 30 + 2 = 32.
The total number of protons on the right side is: 30 + 2 = 32.
c) The total number of electrons on the left side is: 30 + 0 = 30.
The total number of electrons on the right side is: 28 + 2 = 30.
d) Zn(s) is oxidized because it loses electrons (from 30 to 28) and its number increases.