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Chapter 18: Problem 51

In which direction do electrons flow in a galvanic cell, from anode to cathode or vice versa?

Short Answer

Expert verified
In a galvanic cell, electrons flow from the anode to the cathode through the external conductor as a result of oxidation occurring at the anode and reduction occurring at the cathode.

Step by step solution

01

Understanding a Galvanic Cell

A galvanic cell is an electrochemical cell that uses a spontaneous redox (reduction-oxidation) reaction to generate electrical energy. It consists of two electrodes, the anode and the cathode, submerged in an electrolyte solution. The anode is where oxidation occurs, while the cathode is where reduction takes place. These reactions involve the transfer of electrons, and this electron flow is the basis of the electric current.
02

Oxidation at the Anode

At the anode, a chemical species loses one or more electrons in an oxidation half-reaction. These released electrons are then free to move from the anode to another part of the cell. The loss of electrons by a substance is called oxidation.
03

Reduction at the Cathode

At the cathode, another chemical species gains one or more electrons in a reduction half-reaction. The electrons that were released by the anode (through the oxidation process) move through an external conductor (such as a wire) to reach the cathode, where they are accepted by the chemical species undergoing reduction.
04

Electron Flow Direction

In a galvanic cell, electrons are released at the anode (through oxidation) and are accepted at the cathode (through reduction). Therefore, electrons flow from the anode to the cathode through the external conductor. This is the direction that is being asked for in the given exercise. To conclude, in a galvanic cell, electrons flow from the anode to the cathode.

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Key Concepts

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

Electrochemical Cells
Electrochemical cells are vital components used to convert chemical energy into electrical energy. They consist of two main types: galvanic cells, which generate electricity, and electrolytic cells, which consume electricity for chemical reactions. Each electrochemical cell has two electrodes immersed in an electrolyte solution. The electrodes are termed the anode and cathode. Electrons move between these electrodes, driving important chemical processes. By understanding these cells, we learn how devices such as batteries power our everyday tools and appliances.
Electron Flow
The movement of electrons is crucial in an electrochemical cell. This flow creates the current that powers devices. In a galvanic cell, electrons flow spontaneously due to a redox reaction.
They start at the anode, travel through an external wire, and end at the cathode. During this journey, the electrons do work, such as lighting a bulb or running a motor.
  • Anode: source of electrons
  • Cathode: destination for electrons
This directional flow from anode to cathode is consistent in galvanic cells, assisting in the smooth production of electrical energy.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are at the heart of a galvanic cell鈥檚 operation. They involve the transfer of electrons between chemical species.
In these reactions, oxidation occurs when a molecule loses electrons. On the opposite side, reduction is the gain of electrons.
  • Oxidation at the anode: loss of electrons
  • Reduction at the cathode: gain of electrons
These reactions are linked, as electrons lost in oxidation are those gained in reduction. Understanding redox helps clarify why electrons flow the way they do in galvanic cells.

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Most popular questions from this chapter

For each of the following oxidation-reduction reactions of metals with nonmetals, identify which element is oxidized and which is reduced. a. \(4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s)\) b. \(\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{H}_{2}(g)\) c. \(2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g)\) d. \(3 \mathrm{Mg}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}(s)\)

Why must the number of electrons lost in the oxidation equal the number of electrons gained in the reduction? Is it possible to have "leftover" electrons in a reaction?

For each of the following unbalanced oxidation-reduction chemical equations, balance the equation by inspection, and identify which species is the reducing agent. a. \(\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{Al}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{AlCl}_{3}(s)\) c. \(\mathrm{Mg}(s)+\mathrm{P}_{4}(s) \rightarrow \mathrm{Mg}_{3} \mathrm{P}_{2}(s)\)

In each of the following reactions, identify which element is oxidized and which is reduced by assigning oxidation states. a. \(\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\) b. \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)\) c. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{2}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\)

Although magnesium metal does not react with water at room temperature, it does react vigorously with steam at higher temperatures, releasing elemental hydrogen gas from the water. $$ \mathrm{Mg}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g) $$ Identify which element is being oxidized and which is being reduced.
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