Question

A piece of copper metal is dipped into an aqueous solution of AgNO3. Which net ionic equation describes the reaction that occurs? A. Cu(s) + 2Ag+(aq) + 2NO3–(aq) → 2AgNO3(s) + Cu2+(aq) B. 2Ag+(aq) + 2NO3–(aq) → 2AgNO3(s) C. Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq) D. 2NO3–(aq) + Cu2+(aq) → Cu(s) + 2Ag(s)

Answer 1

Answer:

$Cu(s)+2Ag^+(aq) \rightarrow 2Ag(s)+Cu^{2+}(aq)$

Explanation:

When copper metal is dipped into an aqueous solution of $AgNO_3$, copper nitrate ($Cu(NO_3)_2$) is formed. This is type of displacement reaction because copper metal displaces silver from its salt. As only one ion is being displaced, therefore, it is a single displacement reaction.

Copper displaces silver as it is more reactive than silver. This can be confirmed by position of copper and silver in reactivity series. Elements above in the reactive series displaces elements present below from their salts.

The molecular equation is as follows:

$Cu(s)+2AgNO_3(aq) \rightarrow 2Ag(s)+Cu(NO_3)_2(aq)$

Now, write the ionic equation as follows:

$Cu(s)+2Ag^+(aq)+2NO_3^-(aq) \rightarrow 2Ag(s)+Cu^{2+}(aq)+2(NO_3)(aq)$

Cancel the common terms,

$Cu(s)+2Ag^+(aq) \rightarrow 2Ag(s)+Cu^{2+}(aq)$

Answer 2

Answer:

Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq) (option C)

Explanation:

Step 1: the unbalanced equation

Cu(s) + AgNO3(ag) → Ag + Cu(NO3)2

Step 2: The balanced equation

Cu(s) + AgNO3(ag) → Ag + Cu(NO3)2

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 on both sides we have to multiply AgNO3, on the left side, by 2

Cu(s) + 2AgNO3(ag) → Ag + Cu(NO3)2

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

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

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:

Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq) (option C)