Write the chemical equation for this process: ""Copper metal is placed in a solution of silver nitrate. This produces silver metal and copper nitrate solution.""

Cu(s) + 2AgNO₃(aq)  → 2Ag (s) +  Cu(NO₃)₂ (aq)

Explanation:

We must identify the elements of our reaction:

Cu as a reactant

AgNO₃ as another reactant

When both are mixed they make a reaction where there are formed Ag and Cu(NO₃)₂.

This is a redox reaction where:

Cu is reduced ⇒  Cu → Cu²⁺  +  2e⁻

Ag⁺ is oxidized ⇒ Ag⁺ + 1e⁻ → Ag

To balance the reaction, we multiply the second half reaction by 2(x2)

Cu → Cu²⁺  +  2e⁻

(Ag⁺ + 1e⁻ → Ag) .2  ⇒  2Ag⁺ + 2e⁻ → 2Ag

When we sum both, we cancel the electrons:

Cu  +  2Ag⁺ + 2e⁻  →  2Ag  +  Cu²⁺  +  2e⁻

Balanced reaction is:

Cu(s) + 2AgNO₃(aq)  → 2Ag (s) +  Cu(NO₃)₂ (aq)

Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq)

Explanation:

Step 1: data given

Copper metal = Cu(s)

solution of silver nitrate = AgNO3(aq)

silver metal = Ag(s)

copper nitrate solution = Cu(NO3)2(aq)

Step 2: The unbalanced equation

Cu(s) + AgNO3(aq) → Ag(s) + Cu(NO3)2(aq)

Step 3: Balancing the equation

Cu(s) + AgNO3(aq) → Ag(s) + Cu(NO3)2(aq)

On the left side we have 1x NO3 (in AgNO3), on the right side we have 2x NO3 (in Cu(NO3)2). To balance the amount of NO3 we have to multiply AgNO3 on the left side by 2.

Cu(s) + 2AgNO3(aq) → Ag(s) + Cu(NO3)2(aq)

On the left side we have 2x Ag (in 2AgNO3), on the right side we have 1x Ag. To balance the amount of Ag we have to multiply Ag on the right side by 2. Now the equation is balanced.

Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq)

The final result is silver metal precipitating out and a copper II nitrate solution remaining.

It's a displacement reaction as well as redox reaction.

Oxidation number of Cu changes from 0 to 2 so it's oxidized.

Oxidation number of Ag changes from 1 to 0 so it's reduced.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be