Calculate ℰ^∘ values for the galvanic cells in Exercise 25 ....

Name: Problem 27, Chapter 17: Chemistry
Uploaded: 2024-01-04T05:06:41-08:00
Duration: 4 min 40 s
Channel: Dominique Jan Tan
Description: Calculate ℰ^∘ values for the galvanic cells in Exercise 25 .

Key Concepts

Standard Cell Potential Calculation

The standard cell potential is derived from the difference between the standard reduction potentials of the cathode and the anode. It is calculated using the formula E°cell = E°cathode ? E°anode, where the resulting positive value indicates a spontaneous cell reaction. This calculation is fundamental in assessing the driving force behind the galvanic cell's redox process.

Galvanic Cell

A galvanic cell, also known as a voltaic cell, is an electrochemical device that converts chemical energy into electrical energy through spontaneous redox reactions. It consists of two half-cells connected by a conductive medium, where one acts as the anode (site of oxidation) and the other as the cathode (site of reduction), with a salt bridge or porous barrier ensuring ionic conductivity and maintaining charge balance.

Standard Electrode Potential

Standard electrode potential refers to the potential difference between a given half-cell and the standard hydrogen electrode under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It indicates the intrinsic tendency of a species to be reduced and serves as a crucial parameter in predicting the direction of electron flow in redox reactions.

Transcript

00:01 The standard cell potential is computed as the difference between the reduction potential of the species that undergoes reduction in the cathode and the reduction potential of the species that undergoes oxidation in the anode.

00:25 So here we have two reactions which we want to find the cell potential of.

00:32 So first one involves cr3 +, reacting with cl2 to form cr2o7 2 - and cl-.

00:44 So our first task here is to identify the redox or that is the reduction and oxidation half reactions.

00:51 And we do that by separating the species into two, separating the equation into two half reactions by collecting common species.

01:01 So the first one will be cr3 +, with cr2o7 2 - and the second one is cl2 with cl-.

01:10 And then we will just balance this.

01:13 So to balance, we have to balance the number of atoms and then notice that in the reactant side or the product side, and we have the first reaction, we have seven oxygen, but we don't have any in the reactant side.

01:27 So we have to apply 7h2o on that side.

01:33 And since introducing 7h2o also introduces 14 hydrogen, we will put 14h +, on the other side.

01:41 Finally, we just balance the number of electrons.

01:44 We need to put six on the right side and then two electrons on the left side of the second equation.

01:52 The second equation, the first equation shows the loss of electrons, so this is oxidation...

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