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Calculate the standard emf of a cell that uses the $\mathrm{Mg} / \mathrm{Mg}^{2+}$ and $\mathrm{Cu} / \mathrm{Cu}^{2+}$ half-cell reactions at $25^{\circ} \mathrm{C}$ Write the equation for the cell reaction that occurs under standard-state conditions.

$$2.71 V$$

Chemistry 102

Chapter 18

Electrochemistry

Carleton College

Drexel University

Lectures

00:44

In chemistry, electrochemistry is a branch of physical chemistry that studies chemical reactions involving electrical energy, either directly or indirectly. The name derives from the Greek words for amber (electron) and chemistry (chemeia). Electrochemistry deals with the electron transfer reactions that take place at the interface of two electrodes in contact with an electrolyte. The study of these reactions is associated with the study of electrolysis. The process of electrolysis is used to generate a potential difference (voltage) between two electrodes, which is used to drive an electrical current through the electrolytic solution.

03:11

In chemistry, a redox reaction is a chemical reaction in which the oxidation states of the atoms of the participating elements are changed. Any such reaction involves both a reduction process and a complementary oxidation process, two key concepts involved with electron transfer processes. A redox reaction may be written with either an oxidation sign or a reduction sign.

02:27

Calculate the standard emf…

02:46

02:12

02:28

in this problem, we need to calculate the value of the overall cell potential for a cell that contains these 2/2 reactions. We're told that we have M G and M G two plus as well a see you and see you two plus. So that's how we know that these are the 2/2 reactions that are occurring within the cell. And we can look up the standard reduction potentials for both of these half reactions in the table. And now we know that based on the standard reduction potentials that the half reaction with the larger value for the standard reduction potential will be the one that undergoes reduction. Since it has a higher potential to be reduced and therefore the other reaction will be the one that undergoes oxidation. You can see that since 0.34 is greater than negative 2.37 that the bottom hack reaction will be the one that undergoes reduction. So that would be see you two plus gaining two electrons to be reduced to solid copper, and so the top half reaction must be the one that undergoes oxidation. So, in order for magnesium solid to be oxidized. We have to reverse this reaction as well as the sign for its standard reduction potential. After we do that, we added to the reduction reaction force EU two plus to see you. When we write that out. We against said that See you two plus is being reduced in magnesium solid is being oxidized and we flipped the sign to make this now a positive 2.37 for that oxidation. And now we add these together to get the overall so reaction as well as a value for the standard cell potential. You see that we have two electrons cancelling out on each side. So the overall reaction becomes see you two plus a qui ists, a solid magnesium going to M G two plus equi ists plus solid copper. And now, to find the overall cell potential, it's danger conditions. We add those two potentials together So that 0.34 plus 2.37 which comes out to 2.71 volts

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