Asked by ChemGal
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: _____________
(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: _____________
Answers
Answered by
DrBob222
Evidence a change occurred in the test tube: <b>What did you observe?</b>
Balanced eqn: <b> Pb(s) + Cu(NO3)2(aq) ==> Pb(NO3)2(aq) + Cu(s)</b>
Total ionic eqn: <b>
Pb + Cu^2+ + 2[NO3]^- ==> Pb^2+ + 2[NO3]^- + Cu</b>
Net ionic eqn: <b> Remove the spectator ion (the NO3^-). What's left is the net ionic equation.</b>
More active metal: <b> Pb replaced the Cu ion so Pb is the more active metal</b>
Element Oxidized: <b>Which metal (Pb or Cu^2+ lost electrons? The loss of electrons is oxidation.</b>
Element Reduced: <b>The metal not oxidized is the one reduced.</b>
Spectator ion: <b> see above.</b>
Balanced eqn: <b> Pb(s) + Cu(NO3)2(aq) ==> Pb(NO3)2(aq) + Cu(s)</b>
Total ionic eqn: <b>
Pb + Cu^2+ + 2[NO3]^- ==> Pb^2+ + 2[NO3]^- + Cu</b>
Net ionic eqn: <b> Remove the spectator ion (the NO3^-). What's left is the net ionic equation.</b>
More active metal: <b> Pb replaced the Cu ion so Pb is the more active metal</b>
Element Oxidized: <b>Which metal (Pb or Cu^2+ lost electrons? The loss of electrons is oxidation.</b>
Element Reduced: <b>The metal not oxidized is the one reduced.</b>
Spectator ion: <b> see above.</b>
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