A voltaic cell that uses the reaction PdCl 4 2- (aq) + Cd(s) → Pd(s) + 4 Cl - (aq) + Cd 2+ (aq) has a measured standard cell potential of +1.03 V. (a) Write the two half-cell reactions.

Welcome back everyone. We have a certain voltaic cell with a cell potential of plus 1.60V works on an overall reaction of Tetra Bromo or ion with chromium solid to produce solid gold and four bromide and ions and three chrome eight cat ions provide two half cell reactions that are involved. So our first step is to separate these or this entire reaction into two half reactions are first half reaction is going to consist of our story A U B R four minus where we form our solid gold and our formals of our bromide and ion. And our second equation is going to be our solid chromium metal which forms our chrome eight Catalon, where we recognize that this Catalan will gain three electrons to form our solid chromium. Our next step is to make sure our elements are balanced. So looking at both sides of our equation for both of our half reactions, we only have one mole of everything as well as four moles of our booming atoms. So for the first half reaction, everything is balanced. And for the second thing or sorry for the second half reaction, all of our atoms are balanced. So let's confirm that our atoms are balanced in both equations. So what we want to recognize next is that our next step is to balance out our charges. So looking at each of our re agents for the first half reaction, we would recognize that we have an overall oxidation state being minus one for our tetra bro majority where on the product side we have gold in its standard state which would have an oxidation state of zero. And our bromide, an ion which has an oxidation number of minus one times the coefficient four, which would give us a charge of minus four. So we would say its oxidation state is minus four here, meaning that we do not have a balanced oxidation state here. So we're going to have to add electrons to our reactive side of this half reaction to get a balanced charge of minus four on this side as well. So we would add specifically three electrons to this minus one charge here to give us a total charge of negative four on the reactive side. To balance out with our minus four charge on the product side since we know that gold in its standard state will have an oxidation state of zero. Moving on to our second half reaction, we recognize that chromium in its standard state as a metal will have an oxidation state of zero. But on the product side we see that it has a plus three charge. So to balance out the charge here, we're going to add three electrons to our product side here so that it can cancel out this plus three charge leaving us with oxidation or sorry, charges of zero on both sides of the equation here. Now that we have our charges balanced. We want to label the types of reactions we have occurring now because we added electrons to the react inside. So electrons were technically gained here in our first half reaction. We are going to recognize this first half reaction as our reduction half reaction. And moving on to our second half reaction. We have electrons that were lost on the product side here, meaning that this is going to be our oxidation reaction or our oxidation half reaction specifically. And so for our final answers, we have are two half reactions are first being our oxidation half reaction and our second half reaction being our read reduction half reaction here for our second final answer. So it's highlighted in yellow represents our two half cell reactions that are involved. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Ch.20 - Electrochemistry

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Brown 15th Edition

Ch.20 - Electrochemistry

Problem 40a

Chapter 20, Problem 40a

A voltaic cell that uses the reaction PdCl₄^2-(aq) + Cd(s) → Pd(s) + 4 Cl^-(aq) + Cd²⁺(aq) has a measured standard cell potential of +1.03 V. (a) Write the two half-cell reactions.

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Identify the oxidation and reduction components of the overall reaction. In this case, PdCl4^2- is reduced to Pd, and Cd is oxidized to Cd^2+.

Write the reduction half-reaction. Since PdCl4^2- is reduced to Pd, the palladium ion gains electrons. The equation will show the reduction of PdCl4^2- to Pd with the release of Cl^- ions.

Balance the reduction half-reaction for atoms other than hydrogen and oxygen. Ensure that the number of Pd and Cl atoms are equal on both sides of the equation.

Write the oxidation half-reaction. Since Cd is oxidized to Cd^2+, the cadmium metal loses electrons. The equation will show Cd transforming into Cd^2+.

Balance the oxidation half-reaction for atoms and charge. Ensure that the number of Cd atoms and the overall charge are equal on both sides of the equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Half-Cell Reactions

Half-cell reactions represent the oxidation and reduction processes occurring in a voltaic cell. Each half-cell reaction shows the transfer of electrons, with one species being oxidized (losing electrons) and another being reduced (gaining electrons). In the given reaction, identifying the species that undergo oxidation and reduction is essential for writing the half-cell reactions.

Standard Cell Potential

The standard cell potential (E°) is the measure of the voltage produced by a voltaic cell under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It indicates the driving force behind the electrochemical reaction, with positive values suggesting a spontaneous reaction. Understanding how to interpret and calculate standard cell potentials is crucial for analyzing the overall cell reaction.

Electrochemical Series

The electrochemical series is a list of standard electrode potentials for various half-reactions, arranged from the most positive to the most negative. This series helps predict the direction of electron flow in a voltaic cell and the feasibility of reactions. By comparing the standard potentials of the half-reactions involved, one can determine which species will be oxidized and which will be reduced in the cell.