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Chapter 18: Problem 69

Write balanced equations for the electrode and overall cell reactions in the following galvanic cells. Sketch each cell, labeling the anode and cathode and showing the direction of electron and ion flow. (a) \(\mathrm{Mn}(s)\left|\mathrm{Mn}^{2+}(a q) \| \mathrm{Pb}^{2+}(a q)\right| \mathrm{Pb}(s)\) (b) \(\operatorname{Pt}(s)\left|\mathrm{H}_{2}(g)\right| \mathrm{H}^{+}(a q) \| \mathrm{Cl}^{-}(a q)|\mathrm{AgCl}(s)| \mathrm{Ag}(s)\)

Short Answer

Expert verified
(a) Mn(s) + Pb虏鈦�(aq) 鈫� Mn虏鈦�(aq) + Pb(s) (b) H鈧�(g) + AgCl(s) 鈫� 2H鈦�(aq) + Cl鈦�(aq) + Ag(s)

Step by step solution

01

Identify the Redox Reactions

For both galvanic cells, determine the half-reactions that occur at the anode and cathode. For cell (a), Mn(s) undergoes oxidation to Mn虏鈦�(aq), and Pb虏鈦�(aq) undergoes reduction to Pb(s). For cell (b), H鈧�(g) undergoes oxidation to H鈦�(aq), and AgCl(s) undergoes reduction to Ag(s) and Cl鈦�(aq).
02

Write the Half-Reaction Equations

(a) For the Mn|Pb cell: oxidation at the anode is Mn(s) 鈫� Mn虏鈦�(aq) + 2e鈦�. Reduction at the cathode is Pb虏鈦�(aq) + 2e鈦� 鈫� Pb(s). (b) For the Pt|Ag cell: oxidation at the anode is H鈧�(g) 鈫� 2H鈦�(aq) + 2e鈦�. Reduction at the cathode is AgCl(s) + e鈦� 鈫� Ag(s) + Cl鈦�(aq).
03

Balance Each Half-Reaction

Ensure each half-equation is balanced in terms of mass and charge. Both half-equations for the Mn|Pb and Pt|Ag cells are already balanced. Each reaction conserves mass and charge.
04

Write the Overall Cell Reaction

Combine the balanced half-reactions for each cell, making sure the electrons cancel out. (a) Mn(s) + Pb虏鈦�(aq) 鈫� Mn虏鈦�(aq) + Pb(s) (b) H鈧�(g) + AgCl(s) 鈫� 2H鈦�(aq) + Cl鈦�(aq) + Ag(s)
05

Sketch and Label the Galvanic Cells

Draw each cell with the appropriate components. Label the anode (where oxidation occurs) and the cathode (where reduction occurs). Show the direction of electron flow from anode to cathode and ion movement through the salt bridge (for (a): from anode to cathode).

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Key Concepts

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

Electrode Reactions
In the fascinating world of galvanic cells, electrode reactions are fundamental processes that convert chemical energy into electrical energy. These reactions occur at the electrodes, which are crucial components of the cell. Essentially, there are two types of electrode reactions to understand: oxidation and reduction.
  • Oxidation: This reaction involves the loss of electrons from a substance. It takes place at the anode of the galvanic cell.
  • Reduction: On the flip side, reduction involves the gain of electrons by a substance. This process occurs at the cathode of the cell.
Each half-reaction plays a significant role. Together, they facilitate the flow of electrons necessary for generating electricity. Keeping these reactions balanced in terms of charge and mass is crucial for the cell to function effectively.
Anode and Cathode
In a galvanic cell, the anode and cathode are the two electrodes where the essential reactions occur. Understanding these terms is key to grasping how galvanic cells work.
  • Anode: This is the electrode where oxidation happens. - Electrons are lost by the anode material. - This side typically has a negative charge in galvanic cells.
  • Cathode: The site of reduction reactions. - Electrons are gained by the cathode material. - It generally has a positive charge due to the incoming electrons.
Remember, the mnemonic "AN OX and a RED CAT" can help: Anode is where Oxidation happens, and Reduction occurs at the Cathode. This makes it easier to recall the roles each electrode plays.
Electron Flow
In galvanic cells, electron flow is fundamental to the process of electricity generation. Understanding how electrons travel allows us to visualize the conversion from chemical to electrical energy. Electrons generated from the oxidation reaction at the anode travel through an external circuit towards the cathode. This movement is driven by the potential difference between the two electrodes. Here's how it works in simple steps:
  • Electrons are released at the anode during the oxidation reaction.
  • These electrons move through the connecting wire, powering any electrical devices placed in the circuit on their path.
  • Eventually, they reach the cathode where they participate in the reduction reaction.
This electron flow creates an electric current, essential for powering devices, and highlights the practical application of chemistry.
Half-Reaction Equations
Half-reaction equations represent the individual oxidation and reduction processes that occur in galvanic cells. Through these equations, we can deeply understand how the overall cell reaction is composed.
  • Oxidation Half-Reaction: Indicates the loss of electrons at the anode. - An example is: \( \text{Mn}(s) \rightarrow \text{Mn}^{2+}(aq) + 2e^- \)
  • Reduction Half-Reaction: Shows the gain of electrons at the cathode. - An example is: \( \text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \)
Balancing these equations is critical for maintaining charge and mass conservation within the cell. Once balanced, combining these half-reactions forms the overall cell reaction, representing the complete electrochemical process powering a galvanic cell.

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Most popular questions from this chapter

The sodium-sulfur battery has molybdenum electrodes with anode and cathode compartments separated by \(\beta\) -alumina, a ceramic through which sodium ions can pass. Because the battery operates at temperatures above \(300^{\circ} \mathrm{C}\), all the reactants and products are present in a molten solution. The cell voltage is about \(2.0 \mathrm{~V}\). (a) What is the cell reaction if the shorthand notation is \(\mathrm{Mo}(s)\left|\mathrm{Na}(s o l n), \mathrm{Na}^{+}(s o l n) \| \mathrm{S}(s o l n), \mathrm{S}^{2-}(s o l n)\right| \mathrm{Mo}(s) ?\) (b) How many kilograms of sodium are consumed when a \(25 \mathrm{~kW}\) sodium- sulfur battery produces current for \(32 \mathrm{~min} ?\)

What is meant by cathodic protection? (a) Steel is coated with a layer of paint. (b) Iron in steel is oxidized to form a protective oxide coating. (c) Steel is coated with zinc because zinc is more easily oxidized than iron. (d) A strip of magnesium is attached to steel because the magnesium is more easily oxidized than iron.

What are the values of \(x\) and \(y\) for the following reaction if \(E^{\circ}=0.91 \mathrm{~V}\) and \(\Delta G^{\circ}=-527 \mathrm{~kJ} ?\) \(2 \mathrm{~A}^{x+}+3 \mathrm{~B} \longrightarrow 2 \mathrm{~A}+3 \mathrm{~B}^{y+}\)

Balance the following net ionic equation by the half-reaction method. The reaction takes place in acidic solution. $$ \mathrm{NO}_{3}^{-}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{NO}(g)+\mathrm{Cu}^{2+}(a q) \quad \text { Unbalanced } $$

Describe galvanic cells that use the following reactions. In each case, write the anode and cathode half-reactions and sketch the experimental setup. Label the anode and cathode, identify the sign of each electrode, and indicate the direction of electron and ion flow. (a) \(\mathrm{Cd}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Cd}^{2+}(a q)+\mathrm{Sn}(s)\) (b) \(2 \mathrm{Al}(s)+3 \mathrm{Cd}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Cd}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+6 \mathrm{Fe}^{2+}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow\) \(2 \mathrm{Cr}^{3+}(a q)+6 \mathrm{Fe}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)
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