Lead + Copper (II) Nitrate -----> ______ + ______
(Name?) (Name?)
Evidence a change occurred in the test tube: _____________
Balanced eqn: ___________________
Total ionic eqn: ___________________
Net ionic eqn: ___________________
More active metal: ___________ Element Oxidized: _______
Element Reduced: ___________________
Spectator ion: _____________
Evidence a change occurred in the test tube: What did you observe?
Balanced eqn: Pb(s) + Cu(NO3)2(aq) ==> Pb(NO3)2(aq) + Cu(s)
Total ionic eqn:
Pb + Cu^2+ + 2[NO3]^- ==> Pb^2+ + 2[NO3]^- + Cu
Net ionic eqn: Remove the spectator ion (the NO3^-). What's left is the net ionic equation.
More active metal: Pb replaced the Cu ion so Pb is the more active metal
Element Oxidized: Which metal (Pb or Cu^2+ lost electrons? The loss of electrons is oxidation.
Element Reduced: The metal not oxidized is the one reduced.
Spectator ion: see above.
Lead + Copper(II) Nitrate → Lead(II) Nitrate + Copper
Evidence a change occurred in the test tube: A color change may occur, as copper(II) nitrate is blue, while lead(II) nitrate is colorless. Additionally, there may be the formation of a solid precipitate.
Balanced equation: Pb + Cu(NO3)2 → Pb(NO3)2 + Cu
Total ionic equation: Pb + Cu2+ + 2NO3- → Pb2+ + 2NO3- + Cu
Net ionic equation: Pb + Cu2+ → Pb2+ + Cu
More active metal: Copper
Element oxidized: Lead
Element reduced: Copper
Spectator ion: NO3-
To determine the products of the reaction between lead and copper (II) nitrate, we need to understand the reaction types and the solubility rules.
The reaction between a metal and a compound can be categorized as a single replacement reaction. In a single replacement reaction, one element replaces another element in a compound.
To find the products, we need to know the oxidation states of the elements involved. Lead has an oxidation state of +2, indicated by its position in the periodic table. Copper in copper (II) nitrate has an oxidation state of +2. This means lead can replace copper in the compound.
Now, let's analyze the solubility of the compounds formed. Generally, nitrates are soluble, unless they are paired with certain metals. Copper (II) nitrate is soluble since it is a nitrate compound. Lead, on the other hand, typically forms insoluble compounds with nitrates. Based on this information, we can conclude that the products of the reaction will be lead nitrate and copper.
Evidence of a change has occurred in the test tube:
A visible change in color could suggest a chemical reaction. If lead nitrate were formed, it would be a white solid precipitate, which is evidence of a chemical reaction taking place.
The balanced equation for the reaction is:
Pb + Cu(NO3)2 -> Pb(NO3)2 + Cu
The total ionic equation for the reaction is:
Pb + Cu^2+ + 2NO3^- -> Pb^2+ + 2NO3^- + Cu
The net ionic equation for the reaction is:
Pb + Cu^2+ -> Pb^2+ + Cu
The more active metal in the reaction is lead because it replaced copper in the compound.
The element oxidized is copper because it lost electrons and went from an oxidation state of +2 to 0.
The element reduced is lead because it gained electrons and went from an oxidation state of 0 to +2.
The spectator ion in the reaction is nitrate (NO3^-), it remains unchanged throughout the reaction.