Make up the equations for the reactions of aluminum a) with selenium; b) with iodine;

Make up the equations for the reactions of aluminum a) with selenium; b) with iodine; c) with phosphoric acid. Specify the oxidizing agent and reducing agent in these reactions

a) Al + Se —> Al2Se3.
We arrange the oxidation states, taking into account the electronegativity of the atoms:
0 0 +3 -2
Al + Se —> Al2Se3.
Let’s write down the elements that have changed the number of atoms and arrange the coefficients:

0 +3
Al – 3e = Al. * 3 aluminum we donate electrons, it is a reducing agent.

0 -2
Se + 2 e = S. * 1 selenium accepts electrons, it is an oxidizing agent.
Let us sum up the left and right sides of the sex reactions separately:
0 0 +3 -2
Al + Se = Al + Se.
Now let’s put the coefficients into the equation:
2Al + 3Se = Al2Se3.

b) Al + I2 —> Al2Se3.
We arrange the oxidation states, taking into account the electronegativity of the atoms:
0 0 + 3-1
Al + I2 —> AlI3.
Let’s write down the elements that have changed the number of atoms and arrange the coefficients:

0 +3
Al – 3e = Al. * 2 aluminum donates electrons to us, it is a reducing agent.

0 -1
I2 + 2 e = 2I. * 3 iodine takes electrons it is an oxidizing agent.
Let us sum up the left and right sides of the sex reactions separately:
0 0 +3 -1
Al + I2 = Al + 2I.
Now let’s put the coefficients into the equation:
2Al + 3I2 = 2AlI3.

c) Al + H3PO4 —> AlPO4 + H2 ↑.
We arrange the oxidation states, taking into account the electronegativity of the atoms:
0 + 1 + 5-2 +3 + 5-2 0
Al + H3PO4 —> AlPO4 + H2 ↑.
Let’s write down the elements that have changed the number of atoms and arrange the coefficients:

0 +3
Al – 3e = Al. * 2 aluminum donates electrons to us, it is a reducing agent.

+1 0
2H + 2 e = H2. * 3 hydrogen accepts electrons it is an oxidizing agent.
Let us sum up the left and right sides of the reaction sex separately:
0 +1 +3 0
Al + 2H = Al + H2.
Now let’s put the coefficients into the equation:
2Al + 2H3PO4 = 2AlPO4 + 3H2 ↑.