Problem 12 - Chemistry Matter and Change

Question

Write the balanced equation for the spontaneous cell reaction that occurs in a cell with these reduction half-reactions. a. \(A g^{+}(a q)+e^{-} \rightarrow A g(s)\) and \(N i^{2+}(a q)+2 e^{-} \rightarrow\) Ni(s) b. \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Mg}(\mathrm{s})\) and \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g})\) c. \(\mathrm{Sn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}(\mathrm{s})\) and \(\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s})\) d. Pbl \(_{2}(s)+2 e^{-} \rightarrow P b(s)+21-(a q)\) and \(P t^{2+}(a q)+2 e^{-} \rightarrow P t(s)\)

Step-by-Step Solution

Answer

Example a: \( 2Ag^+ + Ni \rightarrow 2Ag + Ni^{2+} \) Example b: \( Mg + 2H^+ \rightarrow Mg^{2+} + H_2 \) Example c: \( 3Sn^{2+} + 2Fe \rightarrow 3Sn + 2Fe^{3+} \) Example d: \( Pb + 2I^- + Pt^{2+} \rightarrow Pbl_2 + Pt \)

Step 1: Determine Half-Reactions for Example (a)

Given the half-reactions: 1. Reduction: \( Ag^+(aq) + e^- \rightarrow Ag(s) \)2. Reduction: \( Ni^{2+}(aq) + 2e^- \rightarrow Ni(s) \)Identify which needs to be reversed to find the oxidation reaction. Nickel has a lower reduction potential and typically oxidizes.

Step 2: Reverse the Appropriate Half-Reaction for Example (a)

Reverse the nickel reduction reaction to show oxidation:- Oxidation: \( Ni(s) \rightarrow Ni^{2+}(aq) + 2e^- \)

Step 3: Balance and Combine Half-Reactions for Example (a)

Balance the electrons transferred:- Multiply the silver half-reaction by 2: \( 2Ag^+(aq) + 2e^- \rightarrow 2Ag(s) \)- Combined balanced equation: \( 2Ag^+(aq) + Ni(s) \rightarrow 2Ag(s) + Ni^{2+}(aq) \)

Step 4: Determine and Balance Example (b)

Given half-reactions:1. \( Mg^{2+}(aq) + 2e^- \rightarrow Mg(s) \)2. \( 2H^+(aq) + 2e^- \rightarrow H_2(g) \)Magnesium will oxidize as it has a greater tendency to lose electrons:- Reverse magnesium reaction: \( Mg(s) \rightarrow Mg^{2+}(aq) + 2e^- \)Combine without adjustment (electrons already balanced):- \( Mg(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2(g) \)

Step 5: Determine and Balance Example (c)

Given half-reactions:1. \( Sn^{2+}(aq) + 2e^- \rightarrow Sn(s) \)2. \( Fe^{3+}(aq) + 3e^- \rightarrow Fe(s) \)Reverse tin reaction for oxidation; balance electrons:- Multiply tin reaction for 3 electrons: \( 3Sn^{2+}(aq) + 6e^- \rightarrow 3Sn(s) \)- Multiply iron for 2 iron atoms: \( 2Fe^{3+}(aq) + 6e^- \rightarrow 2Fe(s) \)Combine:- \( 3Sn^{2+}(aq) + 2Fe(s) \rightarrow 3Sn(s) + 2Fe^{3+}(aq) \)

Step 6: Determine and Balance Example (d)

Given half-reactions:1. \( Pbl_2(s) + 2e^- \rightarrow Pb(s) + 2I^-(aq) \)2. \( Pt^{2+}(aq) + 2e^- \rightarrow Pt(s) \)Platinum has a higher reduction potential, so Reverse lead (II) iodide reaction for oxidation:- \( Pb(s) + 2I^-(aq) \rightarrow Pbl_2(s) + 2e^- \)Equations combine directly:- \( Pb(s) + 2I^-(aq) + Pt^{2+}(aq) \rightarrow Pbl_2(s) + Pt(s) \)

Key Concepts

Reduction Reactions

Reduction reactions are processes in which a substance gains electrons. This gain causes a decrease in the oxidation state of the substance. In electrochemical reactions, half-reactions help in identifying which species is being reduced. For example:
\( Ag^+(aq) + e^- \rightarrow Ag(s) \) shows how silver ions gain an electron to form solid silver, effectively undergoing reduction. This is a key part of electrochemistry, where the reduced species receives electrons that have travelled through an external circuit.

Oxidation Reactions

Oxidation reactions are the opposite of reduction. Here, a species loses electrons, resulting in an increase in its oxidation state. This process is central to balancing redox reactions since oxidation must be paired with a reduction. For instance, reversing the reduction half-reaction of nickel in exercise (a) gives us:
\( Ni(s) \rightarrow Ni^{2+}(aq) + 2e^- \)
This indicates that nickel is oxidized as it loses electrons, contributing to the overall chemical transformation.

Half-Reactions

Half-reactions are vital for understanding redox processes because they separately highlight the reduction and oxidation aspects. Each half-reaction represents either a gain or loss of electrons, crucial for balancing redox equations. By examining half-reactions, such as:
\( Mg^{2+}(aq) + 2e^- \rightarrow Mg(s) \) for reduction and its oxidation counterpart:
\( Mg(s) \rightarrow Mg^{2+}(aq) + 2e^- \),
students can balance electrons and ensure that the electron transfer is equal, producing the net ionic equation.

Chemical Reactions in Electrochemistry

In electrochemistry, chemical reactions involve electron transfer through redox processes, enabling the conversion of chemical energy into electricity and vice versa. Essential concepts here include identifying which half-reactions occur at the anode and cathode. Electrochemical cells typically demonstrate these reactions where one half-reaction at the anode involves oxidation and the other at the cathode involves reduction. By writing balanced equations like:
\( Pb(s) + 2I^-(aq) + Pt^{2+}(aq) \rightarrow Pbl_2(s) + Pt(s) \),
students can visualize how electrons are exchanged during electrochemical reactions, underpinning applications like batteries and electrolysis.