Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells: (a) \(\mathrm{Ni}(\mathrm{s})\left|\mathrm{Ni}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}(\mathrm{s})\) (b) \(\mathrm{C}(\mathrm{gr})\left|\mathrm{H}_{2}(\mathrm{~g}) / \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})\right| \mathrm{Cl}_{2}(\mathrm{~g}) \mid \mathrm{Pt}(\mathrm{s})\) (c) \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ce}^{4+}(\mathrm{aq}), \mathrm{Ce}^{3+}(\mathrm{aq})\right| \operatorname{Pt}(\mathrm{s})\) (d) \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{O}_{2}(\mathrm{~g})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{OH}^{-}(\mathrm{aq})\left|\mathrm{O}_{2}(\mathrm{~g})\right| \mathrm{Pt}(\mathrm{s})\) (c) \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{Sn}^{4+}(\mathrm{aq}), \mathrm{Sn}^{2+}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})\right| \mathrm{Hg}_{2} \mathrm{Cl}{ }_{2}\) (s) \(\mid \mathrm{Hg}(\mathrm{l})\)

Step by step solution

01

Identify Anode Half-Reaction for (a)

At the anode, oxidation takes place. Write the oxidation half-reaction for the element being oxidized: Nickel (Ni) gets oxidized from Ni(s) to Ni^{2+}(aq), giving off 2 electrons: \( \mathrm{Ni(s)} \rightarrow \mathrm{Ni^{2+}(aq)} + 2\mathrm{e^{-}} \).

02

Identify Cathode Half-Reaction for (a)

At the cathode, reduction takes place. Write the reduction half-reaction for the element being reduced: Silver ion (Ag+) gains an electron to become silver metal (Ag): \( \mathrm{Ag^{+}(aq)} + \mathrm{e^{-}} \rightarrow \mathrm{Ag(s)} \).

03

Balance and Combine Half-Reactions for (a)

Balance the electrons in both half-reactions and combine them to get the full balanced redox reaction for the cell: Write the balanced overall equation for the cell reaction: \( \mathrm{Ni(s)} + 2\mathrm{Ag^{+}(aq)} \rightarrow \mathrm{Ni^{2+}(aq)} + 2\mathrm{Ag(s)} \).

04

Identify Anode Half-Reaction for (b)

At the anode, oxidation of hydrogen gas (H2) occurs to form 2 protons (H+) and 2 electrons: \( 2\mathrm{H_2(g)} \rightarrow 4\mathrm{H^{+}(aq)} + 4\mathrm{e^{-}} \).

05

Identify Cathode Half-Reaction for (b)

At the cathode, reduction of chlorine occurs with chloride ions (Cl-) gaining electrons to form chlorine gas (Cl2): \( 2\mathrm{Cl^{-}(aq)} + 2\mathrm{e^{-}} \rightarrow \mathrm{Cl_{2}(g)} \).

06

Balance and Combine Half-Reactions for (b)

Combine the two half-reactions ensuring that the number of electrons lost in oxidation matches the number gained in reduction: \( 2\mathrm{H_2(g)} + 2\mathrm{Cl^{-}(aq)} \rightarrow 4\mathrm{H^{+}(aq)} + \mathrm{Cl_{2}(g)} \).

07

Identify Anode Half-Reaction for (c)

Copper (Cu) is oxidized to Cu^{2+}: \( \mathrm{Cu(s)} \rightarrow \mathrm{Cu^{2+}(aq)} + 2\mathrm{e^{-}} \).

08

Identify Cathode Half-Reaction for (c)

Cerium (Ce) is reduced from Ce^{4+} to Ce^{3+}: \( \mathrm{Ce^{4+}(aq)} + \mathrm{e^{-}} \rightarrow \mathrm{Ce^{3+}(aq)} \).

09

Balance and Combine Half-Reactions for (c)

Balance the electrons and combine the half reactions: \( \mathrm{Cu(s)} + 2\mathrm{Ce^{4+}(aq)} \rightarrow \mathrm{Cu^{2+}(aq)} + 2\mathrm{Ce^{3+}(aq)} \).

10

Identify Half-Reactions for (d)

Write the half-reactions for oxygen at the anode and cathode. Since the same reaction occurs at both electrodes, the net reaction will essentially be the cancellation of electrons. Write the half-reaction for oxygen: \( \mathrm{O_2(g)} + 4\mathrm{H^{+}(aq)} + 4\mathrm{e^{-}} \rightarrow 2\mathrm{H_2O(l)} \).

11

Balance and Combine Half-Reactions for (d)

With both anode and cathode undergoing the same half-reaction and since it's a symmetrical cell, there is no net reaction, because both half-reactions cancel each other out.

12

Identify Anode Half-Reaction for (e)

At the anode, tetravalent tin (Sn^{4+}) gets reduced to divalent tin (Sn^{2+}): \( \mathrm{Sn^{4+}(aq)} \rightarrow \mathrm{Sn^{2+}(aq)} + 2\mathrm{e^{-}} \).

13

Identify Cathode Half-Reaction for (e)

At the cathode, mercuric chloride (Hg2Cl2) gets reduced to mercury (Hg) and chloride ions (Cl-): \( \mathrm{Hg_2Cl_2(s)} + 2e^{-} \rightarrow 2\mathrm{Hg(l)} + 2\mathrm{Cl^{-}(aq)} \).

14

Balance and Combine Half-Reactions for (e)

Adjust the coefficients to ensure electron balance and combine the half-reactions for the net equation: \( \mathrm{Sn^{4+}(aq)} + \mathrm{Hg_2Cl_2(s)} \rightarrow \mathrm{Sn^{2+}(aq)} + 2\mathrm{Hg(l)} + 2\mathrm{Cl^{-}(aq)} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrochemical Cells

An electrochemical cell consists of two half-cells, where chemical reactions produce electrical energy. In a galvanic cell, or voltaic cell, this process occurs spontaneously and is used to harness electricity. These cells include two electrodes (the anode and cathode) and an electrolyte that allows ions to move between them. When a circuit is completed, an electrical current flows as electrons are transferred from the anode, where oxidation (loss of electrons) occurs, to the cathode, where reduction (gain of electrons) takes place.

A key feature of an electrochemical cell is the ability to convert chemical energy into electrical energy, which has vital applications in batteries and power generation.

Oxidation-Reduction (Redox) Reactions

Oxidation-reduction (redox) reactions are chemical processes where electrons are transferred between substances. In these reactions, one substance, the reductant, loses electrons (oxidation) and another substance, the oxidant, gains electrons (reduction).

Understanding redox reactions is crucial because they underpin the functioning of electrochemical cells. In the case of a galvanic cell, we observe redox reactions at the electrodes where different metal elements go through oxidation or reduction, such as Ni becoming Ni²⁺ and Ag⁺ becoming Ag. These changes in oxidation states are the core of harnessing electrical energy from chemical reactions.

Half-Cell Reactions

Half-cell reactions break down the overall redox reaction into two separate processes: the oxidation half-reaction at the anode and the reduction half-reaction at the cathode. This allows for a more straightforward analysis and balancing of the redox process.

In our example, for cell (a), the oxidation half-reaction would be Ni(s) turning into Ni²⁺(aq) by losing electrons, and the reduction half-reaction would be Ag⁺(aq) gaining electrons to become Ag(s). By clearly identifying these half-cell reactions, we ensure the proper flow of electrons and predict the net movement of ions within the cell.

Cell Notation

Cell notation provides a shorthand method to represent the components and reactions within an electrochemical cell. It lists the anode and cathode materials, the states of matter, and separates the different phases with a vertical line, or a double vertical line for the salt bridge.

For instance, in cell (a), the cell notation directly translates to the physical setup, indicating the flow of electrons from the nickel electrode through the external circuit to the silver electrode. This concise notation is crucial for chemists and engineers to quickly understand the design and reactions of an electrochemical cell.

Balancing Chemical Equations

Balancing chemical equations is essential to ensure that the law of conservation of mass is satisfied in a chemical reaction. In the context of electrochemical cells, balancing redox reactions is particularly important as it ascertains the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction.

For example, in cell (a), the two electrons lost by the nickel must be equal to the two electrons gained by the silver ions for the reaction to be balanced. Adjusting coefficients as needed to balance the atoms and charges is a fundamental step in accurately describing the chemistry occurring within an electrochemical cell.