Question

Write the balanced equation for the spontaneous cell reaction that occurs in a cell with these reduction half-reactions. $$ \begin{array}{l}{\text { a. } A g^{+}(a q)+e^{-} \rightarrow A g(s) \text { and } N i^{2+}(a q)+2 e^{-} \rightarrow \text { Nils } )} \\ {\text { b. } M g^{2+}(a q)+2 e^{-} \rightarrow M g(s) \text { and } 2 H^{+}(a q)+2 e^{-} \rightarrow H_{2}(g)}\end{array} $$ $$ \begin{array}{l}{\text { c. } 5 n^{2+}(a q)+2 e^{-} \rightarrow S n(s) \text { and } F e^{3+}(a q)+3 e^{-} \rightarrow F e(s)} \\ {\text { d. } P b l_{2}(s)+2 e^{-} \rightarrow P b(s)+2 l^{-}(a q) \text { and } P t^{2+}(a q)+2 e^{-} \rightarrow P t(s)}\end{array} $$

Short answer

A: a. \( Ni + 2Ag^{+} \rightarrow Ni^{2+} + 2Ag \) b. \( Mg + 2H^{+} \rightarrow Mg^{2+} + H_{2} \) c. \( 3Sn + 2Fe^{3+} \rightarrow 3Sn^{2+} + 2Fe \) d. \( PbI_{2} + Pt \rightarrow Pb + 2I^{-} + Pt^{2+} \)

Step-by-step solution

1. Identifying the reduction and oxidation half-reactions

We will find the oxidation and reduction half-reactions for each reaction. a. Ag+ + e- → Ag (reduction) Ni2+ + 2e- → Ni (reduction) b. Mg2+ + 2e- → Mg (reduction) 2H+ + 2e- → H2 (reduction) c. Sn2+ + 2e- → Sn (reduction) Fe3+ + 3e- → Fe (reduction) d. PbI2 + 2e- → Pb + 2I- (reduction) Pt2+ + 2e- → Pt (reduction)

2. Balancing electrons in the half-reactions

Now, let us balance the electrons in the given half-reactions. a. We will flip the Ni half-reaction as Ni will undergo oxidation. Ag+ + e- → Ag (reduction) Ni → Ni2+ + 2e- (oxidation) b. We will flip the Mg half-reaction as Mg will undergo oxidation. Mg → Mg2+ + 2e- (oxidation) 2H+ + 2e- → H2 (reduction) c. We will flip the Sn half-reaction, as Sn will undergo oxidation. Sn → Sn2+ + 2e- (oxidation) Fe3+ + 3e- → Fe (reduction) d. We will flip the Pt half-reaction, as Pt will undergo oxidation. PbI2 + 2e- → Pb + 2I- (reduction) Pt → Pt2+ + 2e- (oxidation)

3. Write the balanced equation for the spontaneous cell reaction

Now, we balance the number of electrons gained in the reduction half-reaction, with number of electrons lost in the oxidation half-reaction and write the full spontaneous cell reaction. a. 2Ag+ + 2e- → 2Ag (2 electrons reduction) Ni → Ni2+ + 2e- (2 electrons oxidation) Spontaneous reaction: Ni + 2Ag+ → Ni2+ + 2Ag b. Mg → Mg2+ + 2e- (2 electrons oxidation) 2H+ + 2e- → H2 (2 electrons reduction) Spontaneous reaction: Mg + 2H+ → Mg2+ + H2 c. 3Sn → 3Sn2+ + 6e- (6 electrons oxidation) 2Fe3+ + 6e- → 2Fe (6 electrons reduction) Spontaneous reaction: 3Sn + 2Fe3+ → 3Sn2+ + 2Fe d. PbI2 + 2e- → Pb + 2I- (2 electrons reduction) Pt → Pt2+ + 2e- (2 electrons oxidation) Spontaneous reaction: PbI2 + Pt → Pb + 2I- + Pt2+

Key concepts

Oxidation

Oxidation is a key process in redox reactions, crucial to both chemistry and electrochemistry. It involves the loss of electrons by a molecule, atom, or ion. This loss results in an increase in the oxidation state of the chemical species. In simpler terms, whenever a substance loses electrons, it undergoes oxidation.

Consider this as part of a balanced scale, where electrons are transferred from one side to the other:

• An atom or molecule loses electrons = oxidation.
• The oxidation state becomes more positive.

In our example solutions, the reactions were adjusted so that half-reactions represent either oxidation or reduction explicitly. For instance, nickel (Ni) goes from Ni to Ni²⁺ by losing two electrons, signifying oxidation. This change is necessary to balance the equation, as oxidation and reduction must occur together, maintaining the conservation of charge.

Reduction Half-Reactions

Reduction half-reactions are the partner to oxidation in redox reactions. Reduction is the gain of electrons by a molecule, atom, or ion. This process decreases the oxidation state, making it less positive or more negative.

Here’s how to break it down:

• Electrons are gained by an atom or molecule = reduction.
• The oxidation state becomes lower.

In our exercise, silver (Ag⁺) gains an electron to become solid silver (Ag), showing reduction. This pairing of oxidation and reduction half-reactions illustrates how electrons move from one species to another, an essential balance in the chemical reaction. The transferred electrons in reduction half-reactions must match those in oxidation reactions for the equation to balance correctly.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring the Law of Conservation of Mass is followed. This means the total number of atoms and the charge must be the same on both sides of a reaction.To balance, analyze each half-reaction separately:

• Identify how many electrons are transferred in oxidation and reduction.
• Ensure both processes have equal numbers of electrons exchanged.

For instance, the reaction between magnesium (Mg) and hydrogen ions (2H⁺) demonstrates balancing through equalizing electrons gained and lost:\[ \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \] (oxidation)
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \] (reduction)
When balanced correctly, they combine to form the spontaneous reaction:\[ \text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2 \]This balanced approach ensures the reaction accurately represents the mass and charge changes during the process.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical reactions. It explores how redox reactions can produce or consume electricity. Electrical cells and batteries are practical applications of this field.

When discussing electrochemical cells:

• The cell consists of two half-cells, each containing an electrode dipped into an electrolyte solution where redox reactions occur.
• The flow of electrons from one electrode (anode) to the other (cathode) is what produces electric current.

For instance, in a spontaneous cell reaction, nickel acting as the anode gets oxidized, while silver ions at the cathode are reduced. This flow of electrons between these half-cells is harnessed to create an electric current, illustrating electrochemistry at work. Understanding these processes helps in designing batteries and other devices that convert chemical energy into electrical energy.