Question

1. you mix solid zinc with tin (ii) nitrate solution. write the balanced chemical equation list which element is oxidized and which element is reduced.

Explanation:

Step1: Write the un - balanced chemical equation

Zinc (\(Zn\)) reacts with tin (II) nitrate (\(Sn(NO_3)_2\)) to form zinc nitrate (\(Zn(NO_3)_2\)) and tin (\(Sn\)). The un - balanced equation is \(Zn+Sn(NO_3)_2
ightarrow Zn(NO_3)_2 + Sn\)

Step2: Check the number of atoms of each element

• Zinc (\(Zn\)): 1 atom on the left and 1 atom on the right.
• Tin (\(Sn\)): 1 atom on the left and 1 atom on the right.
• Nitrate ion (\(NO_3^-\)): 2 ions on the left (\(Sn(NO_3)_2\)) and 2 ions on the right (\(Zn(NO_3)_2\)).

Since the number of atoms of each element is the same on both sides of the equation, the equation is already balanced.

For oxidation and reduction:

• Oxidation: The process of loss of electrons. Zinc (\(Zn\)) goes from an oxidation state of \(0\) (in \(Zn\)) to \(+ 2\) (in \(Zn(NO_3)_2\)). The oxidation half - reaction is \(Zn

ightarrow Zn^{2+}+2e^-\)

• Reduction: The process of gain of electrons. Tin (\(Sn\)) goes from an oxidation state of \(+2\) (in \(Sn(NO_3)_2\)) to \(0\) (in \(Sn\)). The reduction half - reaction is \(Sn^{2+}+2e^-

ightarrow Sn\)

Answer:

Balanced chemical equation: \(Zn + Sn(NO_3)_2=Zn(NO_3)_2+Sn\)
Oxidized element: Zinc (\(Zn\))
Reduced element: Tin (\(Sn\))