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A voltaic cell is constructed with an \(\mathrm{Sn} / \mathrm{Sn}^{2+}\) half- cell and a \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half-cell. The zinc electrode is negative. (a) Write balanced half-reactions and the overall cell reaction. (b) Diagram the cell, labeling electrodes with their charges and showing the directions of electron flow in the circuit and of cation and anion flow in the salt bridge.

Short Answer

Expert verified
Oxidation: Zn (s) → Zn^{2+} (aq) + 2e^{-}, Reduction: Sn^{2+} (aq) + 2e^{-} → Sn (s), Overall: Zn (s) + Sn^{2+} (aq) → Zn^{2+} (aq) + Sn (s). Draw diagram indicating electron flow from Zn to Sn and ions' direction in the salt bridge.

Step by step solution

01

- Identify Half-Reactions

Determine the oxidation and reduction half-reactions. Zinc electrode is negative, indicating Zn is oxidized while Sn is reduced.
02

- Write Oxidation Half-Reaction

Oxidation at the zinc electrode is as follows: Zn (s) → Zn^{2+} (aq) + 2e^{-}
03

- Write Reduction Half-Reaction

Reduction at the tin electrode is as follows: Sn^{2+} (aq) + 2e^{-} → Sn (s)
04

- Write Overall Cell Reaction

Combine the oxidation and reduction half-reactions to get the overall cell reaction: Zn (s) + Sn^{2+} (aq) → Zn^{2+} (aq) + Sn (s)
05

- Draw Cell Diagram

Draw a two-container setup connected by a salt bridge. Label the zinc electrode as negative (anode) and the tin electrode as positive (cathode). Indicate electron flow from the zinc electrode to the tin electrode through the external circuit. Label the flow of Zn^{2+} ions into the solution and the flow of Sn^{2+} ions out of the solution. Show the flow of anions towards the anode and cations towards the cathode in the salt bridge.

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Key Concepts

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

Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are chemical reactions that involve the transfer of electrons between two substances. One substance loses electrons (oxidation) and another gains electrons (reduction).

In the context of a voltaic cell, like the one given in the exercise, zinc is oxidized, meaning it loses electrons, while tin is reduced, meaning it gains electrons. This can be seen in the half-reactions:

- **Oxidation**: \( \text{Zn (s)} \rightarrow \text{Zn}^{2+} \text{(aq)} + 2e^{-} \)
- **Reduction**: \( \text{Sn}^{2+} \text{(aq)} + 2e^{-} \rightarrow \text{Sn (s)} \)

Together, these half-reactions represent the overall redox process in the cell, resulting in the overall cell reaction: \( \text{Zn (s)} + \text{Sn}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq)} + \text{Sn (s)} \). This exchange of electrons is what generates electrical energy in a voltaic cell.
Electrochemical Cells
Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. There are two main types: voltaic (or galvanic) cells and electrolytic cells.

Voltaic cells, like the one in the exercise, generate spontaneous electrical energy from redox reactions between two different metals or metal compounds. In our example, zinc and tin form the two half-cells of the voltaic cell.

- **Anode**: The electrode where oxidation occurs. It is negative in a voltaic cell. Here, zinc (Zn) is the anode.
- **Cathode**: The electrode where reduction occurs. It is positive in a voltaic cell. Here, tin (Sn) is the cathode.

The flow of electrons from anode to cathode through an external circuit generates electrical energy. The movement of cations towards the cathode and anions towards the anode through a salt bridge maintains charge balance and completes the circuit.
Cell Diagram
A cell diagram is a shorthand representation of an electrochemical cell, showing the arrangement of the different components and their interactions. For the given voltaic cell involving zinc and tin, the cell diagram would look like this:

** Zn (s) | Zn^{2+} (aq) || Sn^{2+} (aq) | Sn (s) **

Here is how the cell diagram is structured:
- The single vertical line (|) represents a phase boundary, such as between a solid electrode and an aqueous solution.
- The double vertical line (||) represents the salt bridge between the two half-cells.

In the cell, electrons flow from the zinc electrode (anode) to the tin electrode (cathode), while the salt bridge allows ions to flow to balance the charges. This process can be visualized in a two-container setup, labeled with:
- The zinc electrode (anode) on the left, labeled negative, with Zn^{2+} ions flowing into the solution as zinc is oxidized.
- The tin electrode (cathode) on the right, labeled positive, with Sn^{2+} ions being reduced and depositing tin metal.
- The external circuit connecting the electrodes and completing the electrical flow path.
- The salt bridge, allowing anionic and cationic flow to maintain electrical neutrality.

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Most popular questions from this chapter

Brass, an alloy of copper and zinc, can be produced by simultaneously electroplating the two metals from a solution containing their \(2+\) ions. If \(65.0 \%\) of the total current is used to plate copper, while \(35.0 \%\) goes to plating zinc, what is the mass percent of copper in the brass?

What is the difference between an active and an inactive electrode? Why are inactive electrodes used? Name two substances commonly used for inactive electrodes.

In Appendix \(D,\) standard electrode potentials range from about \(+3 \mathrm{~V}\) to \(-3 \mathrm{~V}\). Thus, it might seem possible to use a half- cell from each end of this range to construct a cell with a voltage of approximately 6 V. However, most commercial aqueous voltaic cells have \(E^{\circ}\) values of \(1.5-2 \mathrm{~V}\). Why are there no aqueous cells with significantly higher potentials?

Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution: $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{Cr}^{3+}$$

Comparing the standard electrode potentials \(\left(E^{\circ}\right)\) of the Group \(1 \mathrm{~A}(1)\) metals \(\mathrm{Li}, \mathrm{Na},\) and \(\mathrm{K}\) with the negative of their first ionization energies reveals a discrepancy: Ionization process reversed: \(\mathrm{M}^{+}(g)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(g)\) Electrode reaction: $$\mathrm{M}^{+}(a q)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(s)$$ $$\begin{array}{lcc}\text { Metal } & -\text { IE (kJ/mol) } & E^{\circ} \text { (V) } \\\\\hline \text { Li } & -520 & -3.05 \\\\\text { Na } & -496 & -2.71 \\\ \text { K } & -419 & -2.93\end{array}$$ Note that the electrode potentials do not decrease smoothly down the group, whereas the ionization energies do. You might expect that, if it is more difficult to remove an electron from an atom to form a gaseous ion (larger IE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smaller \(E^{\circ}\) ), yet \(\mathrm{Li}^{+}(a q)\) is more difficult to reduce than \(\mathrm{Na}^{+}(a q) .\) Applying Hess's law, use an approach similar to a Born-Haber cycle to break down the process occurring at the electrode into three steps, and label the energy involved in each step. How can you account for the discrepancy?

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