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part b: short answer and calculations (thinking and inquiry and applica…

Question

part b: short answer and calculations (thinking and inquiry and application)

  1. for the following reaction, write (and clearly label) the complete oxidation half - reaction and the complete reduction half - reaction (including the number of electrons transferred in each). also, indicate who the reducing agent is and who the oxidizing agent is.

\ce{mg_{(s)} + 2ag^{+}_{(aq)} -> mg^{2+}_{(aq)} + 2ag_{(s)}}

overall expectations: f2 and f3not yetlevel 4++ 100%level 4+ 95%level 4 85%level 3 75%level 2 65%level 1 55%

Explanation:

Step1: Identify Oxidation Half - Reaction

Oxidation is the loss of electrons. In the reaction \(Mg_{(s)}+2Ag^{+}_{(aq)}
ightarrow Mg^{2 + }_{(aq)}+2Ag_{(s)}\), magnesium (\(Mg\)) goes from an oxidation state of 0 (in \(Mg_{(s)}\)) to + 2 (in \(Mg^{2+}_{(aq)}\)). So the oxidation half - reaction is the process where \(Mg\) loses electrons. The oxidation half - reaction is: \(Mg_{(s)}
ightarrow Mg^{2+}_{(aq)} + 2e^-\). Here, \(Mg\) loses 2 electrons.

Step2: Identify Reduction Half - Reaction

Reduction is the gain of electrons. Silver (\(Ag\)) goes from an oxidation state of + 1 (in \(Ag^{+}_{(aq)}\)) to 0 (in \(Ag_{(s)}\)). For 2 moles of \(Ag^{+}\), the reduction half - reaction is: \(2Ag^{+}_{(aq)}+2e^-
ightarrow 2Ag_{(s)}\). Here, each \(Ag^{+}\) gains 1 electron, and for 2 \(Ag^{+}\) ions, a total of 2 electrons are gained.

Step3: Identify Oxidizing and Reducing Agents

The reducing agent is the substance that is oxidized (loses electrons). Since \(Mg\) is oxidized (loses electrons in the oxidation half - reaction), \(Mg\) is the reducing agent. The oxidizing agent is the substance that is reduced (gains electrons). Since \(Ag^{+}\) is reduced (gains electrons in the reduction half - reaction), \(Ag^{+}\) (or the species containing \(Ag^{+}\), in this case, the \(Ag^{+}_{(aq)}\) from the reactants) is the oxidizing agent.

Answer:

  • Oxidation half - reaction: \(Mg_{(s)}

ightarrow Mg^{2+}_{(aq)} + 2e^-\) (oxidation, 2 electrons lost)

  • Reduction half - reaction: \(2Ag^{+}_{(aq)}+2e^-

ightarrow 2Ag_{(s)}\) (reduction, 2 electrons gained)

  • Reducing agent: \(Mg_{(s)}\) (because it is oxidized, loses electrons)
  • Oxidizing agent: \(Ag^{+}_{(aq)}\) (because it is reduced, gains electrons)