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
Explanation:
There is some info missing. I will add the complete question.
<em>Consider the following oxidation/reduction reaction.</em>
<em> Zn(s) + 2 H⁺(aq) ⇒ Zn²⁺(aq) + H₂(g)</em>
<em>#p⁺ 2</em>
<em>#e⁻ 2</em>
<em>a) Fill in the number of protons and electrons for each product and reactant (two boxes have been filled in for you).</em>
<em>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.</em>
<em>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.</em>
<em>d) Which substance is being oxidized?</em>
<em />
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.