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A voltaic cell is based on the reaction between \(\mathrm{Cu}^{2+}(a q)\) and \(\mathrm{Ni}(s),\) producing \(\mathrm{Cu}(s)\) and \(\mathrm{Ni}^{2+}(a q)\) a. Write the anode and cathode half-reactions. b. Write a balanced cell reaction. c. Draw the cell diagram.

Short Answer

Expert verified
Write the balanced cell reaction and draw the cell diagram. Answer: The anode half-reaction is Ni(s) 鈫 Ni虏鈦(aq) + 2e鈦, and the cathode half-reaction is Cu虏鈦(aq) + 2e鈦 鈫 Cu(s). The balanced cell reaction is Ni(s) + Cu虏鈦(aq) 鈫 Ni虏鈦(aq) + Cu(s). The cell diagram is: Ni (s) | Ni虏鈦 (aq) || Cu虏鈦 (aq) | Cu (s).

Step by step solution

01

1. Identify the oxidation and reduction half-reactions

The overall reaction involves the transfer of electrons between Cu虏鈦 and Ni. Remember that in a redox reaction, one species is reduced (gains electrons) and the other is oxidized (loses electrons). In this case, Cu虏鈦 gains electrons (2) to form Cu(s), while Ni(s) loses two electrons to form Ni虏鈦(aq). Thus, we have: Oxidation half-reaction: Ni(s) 鈫 Ni虏鈦(aq) + 2e鈦 Reduction half-reaction: Cu虏鈦(aq) + 2e鈦 鈫 Cu(s)
02

2. Identify anode and cathode half-reactions

In a voltaic cell, the oxidation half-reaction occurs at the anode and the reduction half-reaction occurs at the cathode. Therefore: Anode half-reaction: Ni(s) 鈫 Ni虏鈦(aq) + 2e鈦 Cathode half-reaction: Cu虏鈦(aq) + 2e鈦 鈫 Cu(s)
03

3. Write a balanced cell reaction

To write the balanced cell reaction, we can combine the anode and cathode half-reactions. Since the number of electrons transferred is the same in both half-reactions, they can be directly added together: Ni(s) 鈫 Ni虏鈦(aq) + 2e鈦 (Anode) Cu虏鈦(aq) + 2e鈦 鈫 Cu(s) (Cathode) 鈥斺斺斺斺斺斺斺斺斺 Balanced cell reaction: Ni(s) + Cu虏鈦(aq) 鈫 Ni虏鈦(aq) + Cu(s)
04

4. Draw the cell diagram

To draw the cell diagram, follow these steps: 1. Write the anode half-reaction on the left, starting with the reactants and then the products: Ni (s) | Ni虏鈦 (aq) 2. Write the cathode half-reaction on the right, starting with the reactants and then the products: Cu虏鈦 (aq) | Cu (s) 3. Separate the two half-reactions with a double vertical line, which represents the salt bridge: || The final cell diagram is: Ni (s) | Ni虏鈦 (aq) || Cu虏鈦 (aq) | Cu (s)

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Key Concepts

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

Redox Reactions
Redox reactions are a fascinating class of chemical reactions where electrons are transferred between substances. The term 'redox' is short for reduction-oxidation, referring to the two key processes involved.
Reduction is the gain of electrons by a molecule, atom, or ion. In contrast, oxidation is the loss of electrons. In any redox reaction, both processes occur simultaneously: as one substance is oxidized, another is reduced.

**Key elements of redox reactions:**
  • **Oxidizing Agent**: The substance that gains electrons and is reduced.
  • **Reducing Agent**: The substance that loses electrons and is oxidized.
  • **Electron Transfer**: Fundamental to all redox reactions. This transfer is what fuels the reaction.
In the provided voltaic cell example, the copper ions (\(\mathrm{Cu}^{2+}\)) gain two electrons to become solid copper (\(\mathrm{Cu(s)}\)) 鈥 this is reduction. Meanwhile, nickel (\(\mathrm{Ni(s)}\)) loses electrons to form nickel ions (\(\mathrm{Ni}^{2+}\)) 鈥 this is oxidation. Both steps contribute to the overall balanced cell reaction.
Anode and Cathode
A voltaic cell, also known as a galvanic cell, depends on two electrodes: the anode and the cathode.
These electrodes facilitate the redox reaction by allowing the flow of electrons through an external circuit.

**Anode Facts:**
  • The anode is the electrode where oxidation occurs. Electrons are lost by the substance here.
  • In the example provided, nickel (\(\mathrm{Ni(s)}\)) acts as the anode. It loses electrons and converts into (\(\mathrm{Ni}^{2+}\)).
  • In a voltaic cell, the anode carries a negative charge because it supplies electrons to the circuit.
**Cathode Facts:**
  • The cathode is the site of reduction where electrons are gained.
  • In this cell, copper ions (\(\mathrm{Cu}^{2+}\)) gain electrons at the cathode to form solid copper (\(\mathrm{Cu(s)}\)).
  • The cathode is positively charged because it receives electrons.
Understanding the role of anodes and cathodes is crucial to grasp how voltaic cells harness chemical energy to produce electricity.
Cell Diagram
A cell diagram succinctly represents the components of a voltaic cell and their interactions.
It provides a snapshot of the species present in each electrode compartment and the flow of electrons.

**How to Read a Cell Diagram:**
  • The diagram begins on the left with the anode, proceeds through the salt bridge, and ends on the right with the cathode.
  • A single vertical line '|' signifies a phase boundary within the same half-cell.
    For instance, in our example, 'Ni(s) | Ni\(^{2+}\)(aq)' indicates the solid nickel is turning into ions.
  • The double vertical line '||' represents the salt bridge that connects the two half-cells, allowing ion flow without mixing the solutions.
  • The right side, 'Cu\(^{2+}\)| Cu(s)', shows the reduction half-reaction.
The cell diagram for our exercise is 'Ni(s) | Ni\(^{2+}\)(aq) || Cu\(^{2+}\)(aq) | Cu(s)', clearly showing the direction of the redox processes and aiding in understanding the voltaic cell's function.

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Most popular questions from this chapter

Which of the following voltaic cell reactions delivers more electrical energy per gram of anode material at \(25^{\circ} \mathrm{C} ?\) $$ \mathrm{Zn}(s)+\mathrm{Ni}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{Zn}(\mathrm{OH})_{2}(s)+\mathrm{Ni}(s) \quad E_{\mathrm{cell}}^{*}=1.50 \mathrm{V}$$ or$$2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ZnO}(s) \quad E_{\mathrm{cell}}^{\circ}=2.08 \mathrm{V}$$

A voltaic cell with a basic aqueous background electrolyte is based on the oxidation of \(\mathrm{Cd}(s)\) to \(\mathrm{Cd}(\mathrm{OH})_{2}(s)\) and the reduction of \(\mathrm{MnO}_{4}^{-}(a q)\) to \(\mathrm{MnO}_{2}(s)\) a. Write half-reactions for the cell's anode and cathode. b. Write a balanced net ionic equation describing the cell reaction. c. Draw the cell diagram.

Describe two advantages of hybrid (gasoline engineelectric motor) power systems over all-electric systems based on fuel cells. Describe two disadvantages.

Select the appropriate half-reactions from Appendix 6 to write net ionic equations describing the reaction between: a. tin and \(A g^{+}\) ions in solution that produces dissolved \(\mathrm{Sn}^{2+}\) ions and silver metal. b. copper and \(\mathrm{O}_{2}\) in an acidic solution that produces \(\mathrm{Cu}^{2+}\) ions. c. solid \(\mathrm{Cr}(\mathrm{OH})_{3}\) and \(\mathrm{O}_{2}\) in a basic solution that produces \(\mathrm{CrO}_{4}^{2-}\) ions.

The anode in an electrochemical cell is defined as the electrode where oxidation takes place. Why is the anode in an electrolytic cell connected to the positive \((+)\) terminal of an external supply, whereas the anode in a voltaic cell battery is connected to the negative \((-)\) terminal?

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