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Define the following terms: anode, cathode, cell voltage, electromotive force, standard reduction potential.

Short Answer

Expert verified
Anode is where oxidation occurs; cathode is where reduction occurs; cell voltage is the potential difference; EMF is the maximum potential difference; standard reduction potential measures tendency to gain electrons.

Step by step solution

01

Anode Definition

An anode is the electrode in an electrochemical cell where oxidation occurs. This means that electrons are given up by the chemical species at the anode. In a galvanic cell, the anode is the negative terminal, because electrons are flowing out of it.
02

Cathode Definition

A cathode is the electrode where reduction occurs. Electrons from the external circuit enter the cathode, reducing the chemical species present at the surface of the cathode. In a galvanic cell, the cathode is the positive terminal.
03

Cell Voltage Explanation

Cell voltage, also known as cell potential, is the measurement of the potential difference between the anode and cathode of an electrochemical cell. It is essentially the force that drives the flow of electrons through the circuit.
04

Electromotive Force Description

Electromotive force (EMF) is the maximum potential difference between the electrodes of a cell when no current flows. EMF is often used to describe the 'driving force' behind the movement of electrons in an electrochemical cell.
05

Standard Reduction Potential Description

Standard reduction potential is the measure of the tendency of a chemical species to acquire electrons and be reduced, measured under standard conditions (25°C, 1 M concentration, 1 atm pressure). It is typically measured in volts and compared against the standard hydrogen electrode, which is assigned a potential of 0 volts.

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Key Concepts

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

Anode
In an electrochemical cell, the anode is a crucial component where oxidation occurs. This is the part of the cell where chemical species lose electrons. You can think of the anode as a source of electrons. In a galvanic cell, which is commonly used for batteries, the anode is typically designated as negative because electrons are generated here and flow out to the rest of the circuit.

The process of oxidation at the anode is essential because it is one-half of the electrochemical reaction happening in the cell. It's the starting point for the flow of electrons which then move to the other electrode, the cathode. Understanding the role of the anode helps in grasping how electricity is produced in various devices.
  • Anode = Oxidation occurs
  • Negative terminal in galvanic cells
  • Source of electrons
Cathode
In contrast to the anode, the cathode is the site of reduction. Here, the chemical species are gaining electrons. In the context of an electrochemical cell, a cathode receives electrons from the external circuit. This characteristic makes the cathode the positive terminal in galvanic cells, where it attracts electrons.

Reduction at the cathode is as important as oxidation at the anode because it completes the circuit. The intake of electrons facilitates the continuation of the chemical reaction that generates electric power. Without the cathode, the flow of electricity would halt. So, in simple terms:
  • Cathode = Reduction occurs
  • Positive terminal in galvanic cells
  • Receives electrons
Cell Voltage
Cell voltage, or cell potential, is a key concept in understanding how much energy is available from an electrochemical cell. It represents the potential difference between the anode and cathode. In other words, it's the voltage, or 'push', that drives electrons from the anode to the cathode through the external circuit.

Cell voltage can directly indicate the effectiveness of a cell to do work. A higher cell voltage means more energy is available for electrical work. It is often measured in volts and can vary depending on the materials used for the anode and cathode. When working with electrochemical cells, understanding the cell voltage can help predict performance.
  • Cell Voltage = Potential difference between terminals
  • Drives electron flow
  • Measured in volts
Electromotive Force
Electromotive force, often abbreviated as EMF, represents the cell's potential to drive electric current under ideal, no-current conditions. It is essentially the maximum voltage a cell can provide. Think of EMF as the starting push that makes electrons move when the circuit is first connected.

It sets the stage for the flow of electrons that occurs once the circuit is closed, even though once the current begins, a small drop in voltage is experienced. Nevertheless, EMF is critical because it gives an idea of the cell's perfect ability to generate energy without any loss. When measuring or discussing cells, knowing the EMF helps understand their maximum efficiency.
  • EMF = Maximum potential difference with no current
  • Initial driving force for electrons
  • Indicator of cell's efficiency
Standard Reduction Potential
Standard reduction potential describes how likely a chemical species is to gain electrons under standard conditions, such as a temperature of 25°C, a concentration of 1 M, and a pressure of 1 atm. It is a crucial measure because it allows comparison between different reactions in terms of their ability to acquire electrons.

The standard hydrogen electrode (SHE) is the benchmark for measuring these potentials, set at 0 volts. Other chemicals are compared against this reference. Understanding standard reduction potentials helps predict the direction of electron flow in electrochemical cells. Higher values suggest a greater tendency to be reduced, making this concept vital for designing efficient cells.
  • Standard Reduction Potential = Electron acquisition tendency
  • Measured under standard conditions
  • Compared against SHE

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Most popular questions from this chapter

The half-reaction at an electrode is: \(\mathrm{Mg}^{2+}(\) molten \()+2 e^{-} \longrightarrow \mathrm{Mg}(s)\) Calculate the number of grams of magnesium that can be produced by supplying \(1.00 \mathrm{~F}\) to the electrode.

Given that: $$ \begin{array}{ll} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)$$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

"Galvanized iron" is steel sheet that has been coated with zinc; "tin" cans are made of steel sheet coated with tin. Discuss the functions of these coatings and the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface of a galvanized iron sheet or a tin can.

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to Fe(II) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

A piece of magnesium ribbon and a copper wire are partially immersed in a \(0.1 M \mathrm{HCl}\) solution in a beaker. The metals are joined externally by another piece of metal wire. Bubbles are seen to evolve at both the \(\mathrm{Mg}\) and Cu surfaces. (a) Write equations representing the reactions occurring at the metals. (b) What visual evidence would you seek to show that Cu is not oxidized to \(\mathrm{Cu}^{2+} ?(\mathrm{c})\) At some stage, \(\mathrm{NaOH}\) solution is added to the beaker to neutralize the HCl acid. Upon further addition of \(\mathrm{NaOH},\) a white precipitate forms. What is it?

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