write the balanced molecular and net iconic equations for the reaction between aluminium metal and silver nitrate. Identify the oxidation and reduction half-reaction

The molecular equation is

3AgNO3 + Al ==> Al(NO3)3 + 3Ag

The net ionic equation is
3Ag^+ + Al ==> Al^3+ + 3Ag

The half reactions are
Ag^+ + e==> Ag
Al ==> Al^3+ + 3e

Oxidation is the loss of electrons; reduction is the gain of electrons.

To write the balanced molecular equation for the reaction between aluminum metal and silver nitrate, we first need to know the formulas of the compounds involved.

The formula for aluminum is Al, and the formula for silver nitrate is AgNO3.

The balanced molecular equation can be written as follows:

2Al + 3AgNO3 ⟶ 3Ag + 2Al(NO3)3

To write the net ionic equation, we need to identify which ions are present in the reaction and which ones undergo a chemical change.

In this reaction, the aluminum ion (Al3+) undergoes oxidation, while the silver ion (Ag+) undergoes reduction.

The balanced net ionic equation is written as follows:

2Al + 6Ag+ + 6NO3- ⟶ 3Ag + 2Al(NO3)3

In this equation, the Al3+ ion loses three electrons (undergoing oxidation) to form aluminum nitrate (Al(NO3)3), while the Ag+ ion gains an electron (undergoing reduction) to form silver (Ag).

To summarize, the oxidation half-reaction is:

2Al ⟶ 2Al3+ + 6e-

And the reduction half-reaction is:

6Ag+ + 6e- ⟶ 6Ag

Sure! Let's start by writing the balanced molecular equation for the reaction between aluminum metal (Al) and silver nitrate (AgNO3):

2 Al + 3 AgNO3 → 2 Al(NO3)3 + 3 Ag

To write the net ionic equation, we need to break down the soluble compounds into their respective ions and include only the species that participate in the reaction. In this case, aluminum nitrate (Al(NO3)3) is soluble and dissociates into its constituent ions, while silver (Ag) is insoluble and remains as a solid:

2 Al(s) + 6 Ag+(aq) + 6 NO3-(aq) → 2 Al3+(aq) + 6 NO3-(aq) + 3 Ag(s)

Since the nitrate ions (NO3-) appear unchanged on both sides of the equation, they can be eliminated, leaving:

2 Al(s) + 6 Ag+(aq) → 2 Al3+(aq) + 3 Ag(s)

Now let's identify the oxidation and reduction half-reactions:

Oxidation half-reaction:
Al(s) → Al3+(aq) + 3e-

Reduction half-reaction:
6 Ag+(aq) + 6e- → 6 Ag(s)

In the oxidation half-reaction, aluminum loses three electrons (oxidation) to form aluminum ions. In the reduction half-reaction, six silver ions gain six electrons (reduction) to form solid silver.

I hope this helps! Let me know if you have any further questions.