Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride and hydrogen gas. Include states. From the equation, which element is oxidized, and which element is reduced?

The balanced chemical equation for the reaction is

Sn(s) + 2HCl(aq) → SnCl₂(aq) + H₂(g)

and the element that oxidized is tin (Sn) while the element that is reduced is hydrogen (H)

The balanced chemical equation for the reaction where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride and hydrogen gas is

Sn(s) + 2HCl(aq) → SnCl₂(aq) + H₂(g)

From the equation,

The element that is oxidized is tin (Sn), because the oxidation state of Sn increased from 0 to +2

The element that is reduced is hydrogen (H), because the oxidation state of H decreased from +1 to 0.

Hence,

The balanced chemical equation for the reaction is

Sn(s) + 2HCl(aq) → SnCl₂(aq) + H₂(g)

and the element that oxidized is tin (Sn) while the element that is reduced is hydrogen (H).

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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