Answer:
1. The balanced equation is given below:
Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)
2a. H is oxidized.
2b. Sn is reduced.
Explanation:
1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).
This is illustrated below:
Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)
There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:
Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)
Now, the equation is balanced
2. Determination of the element that is oxidize and reduced.
This can be obtained as follow:
We shall determine the change in oxidation number of each element.
NOTE:
a. The oxidation number of H is always +1 except in hydrides where it is –1.
b. The oxidation state of Cl is always –1.
Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)
For Tin (Sn):
Sn = 0
SnCl₂ = 0
Sn + 2Cl = 0
Cl = – 1
Sn + 2(–1) = 0
Sn – 2 = 0
Collect like terms
Sn = 0 + 2
Sn = +2
Therefore, the oxidation number of Tin (Sn) changes from 0 to +2
For H:
H = +1
H₂ = 0
The oxidation number of H changes from +1 to 0
For Cl:
Cl is always –1. Therefore no change.
Summary:
Element >>Change in oxidation number
Sn >>>>>>>From 0 to +2
H >>>>>>>>From +1 to 0
Cl >>>>>>>No change
Therefore,
Sn is reduced since its oxidation number increased from 0 to +2.
H is oxidized since it oxidation number reduced from +1 to 0
