When solid copper is placed in an aqueous silver nitrate solution, a reaction occurs, as represented by the equation below.
Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq)
61 State the change in oxidation state of copper in this reaction.
Solution: Let's review some oxidation number rules.
Rules for determining oxidation numbers:
Free, neutral elements have an oxidation number of 0.
Group 1 metals have an oxidation number of +1 (can be found on the Periodic Table)
Group 2 metals have an oxidation number of +2
Oxygen is -2 in all compounds other than peroxide (H2O2)
Hydrogen is +1 unless it is bonded to a metal
The sum of oxidation numbers in a neutral molecule is 0
The sum of oxidation numbers in a charged molecule (ion) is equal to the charge Looking at the equation given, on the left side Cu(s) is on its own and must have an oxidation number of 0. On the right we have Cu(NO3)2. The charge for NO3 is -1. The charge numbers get cross multiplied. Since we see 2 next to NO3 in the formula, the oxidation number of copper was 2+. The change is 0 to +2.
62 Based on Table J, state why Cu(s) reacts spontaneously with Ag+(aq).
Solution: Metals that are higher on table J are more active and can replace other metals in a single replacement reaction. Since Cu is above Ag on table J, it can replace Ag.
63 Write a balanced half-reaction equation to represent the reduction of the silver ions to
silver atoms.
Solution: Let's start with a couple of mnemonics to help us memorize necessary information.
OIL = Oxidation Is Loss of Electrons
RIG = Reduction Is Gain of Electrons
Reduction is gain of electrons. We saw in question 61 that copper went from 0 to +2 which means it lost two electrons and therefore was oxidized. Ag goes from +1 to 0 in the equation (from Ag+1 to Ag). Therefore it gets reduced. Let's write the half reaction equation for it.
Ag+(aq) + e- → Ag(s)
We add the electron on the left side to balance the charge.
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