The cell potential of this electrochemical cell depends on the gold concentration in the cathode half-cell. Pt(s) | H 2 (g, 1.0 atm) | H + (aq, 1.0 M) || Au 3+ (aq, ? M) | Au(s) What is the concentration of Au 3+ in the solution if E cell is 1.22 V?

welcome back Everyone were given the below electrochemical cell and we're told that the cell potential is dependent on our copper concentration which is in our cathode half cell. We're told that the cell potential for the cell is 0.34 volts. And to determine the concentration of copper plus ions for the solution. So we need to write out our half reactions first and our first half reaction is for the oxidation of hydrogen gas. And so we form our hydrogen gas from our hydrogen proton, gaining an electron for our second half reaction, we have the reduction of our copper caddy on, so our copper caddy on gains an electron to form solid copper. Now we know our first half reaction is an oxidation because we have an electron added to the right hand side and we should recall that our oxidation occurs at the node portion of our electoral voltaic cell. Whereas we know that the second half reaction as a reduction because we have an electron added to the react inside. And so we should recall that this is a reduction and will occur at the cathode of our electoral voltaic cell and that is what the prompt tells us. Our reduction occurs at the cathode half cell. Now, according to our cell potential table in our textbooks, the first half reaction has a cell potential equal to a value of zero volts. Whereas our second half reaction has a cell potential value equal to 0.52 volts. Our next step is to make sure whether we can add up these two reactions. And right now we need to see if the first half reaction is balanced. We have two atoms of hydrogen on the react inside and one atom of hydrogen proton on the right hand side. And we need to balance this out by placing a coefficient of two in front of our hydrogen proton. So now that the atoms are balanced. Now looking at our second half reaction, we have one atom of copper on the left hand side and one atom of copper on the product side. So the atoms of copper are balanced now because we added a coefficient of two to our first half reaction, we now have two hydrogen protons that are going to be gaining an electron. And so this means that we actually need to place a coefficient of two in front of our electrons as well so that both of the hydrogen protons are gaining an electron. And so what this means now is that our electrons need to be balanced in order to add up these two half reactions. So to add these two half reactions up, we're going to multiply the second half reaction by two. And in doing so so I'll actually bring down the second the plus sign down here. So our first reaction is still going to remain the same. But our second reaction is now going to be two moles of copper caddy on gaining two electrons to produce two moles of solid copper. And so now we will add this to our first half reaction where we have one mole of hydrogen gas which forms two moles of hydrogen proton and two electrons. So our electrons and atoms are balanced in both of these half reactions. And now we're going to be able to add them together as well as recognize that we can cancel out our electrons because we have two electrons in the reacting side in the first half reaction and two electrons on the product side in the second half reaction. And so now we would have two moles and sorry about that. We have two moles of our copper cat ion added to one mole of hydrogen gas which produces two moles of solid copper plus two moles of our hydrogen proton. And we can agree that according to our reactions, we have the transfer of two electrons. So we would say that n which is our numbers of electrons transferred is equal to two. Our next step in our solution is to go ahead and find out our standard cell potential. So that would be e degree cell And we should recall that we can calculate this by taking the self potential of our cathode subtracted from the cell potential of our a node. So as we stated, our oxidation half reaction occurs at our anodes and had a cell potential of 0V. Whereas our cell potential of 50.52 volts occurs at the reduction half reaction where our cathode is. And so we would say we have 0.52 volts subtracted from our cell potential over a note which is just zero volts. And so this gives us our standard cell potential equal to a value of just 00.52 volts. Our next step is to determine our reaction quotient Q. For the ratio of the ions in our half cell. And we should recall that this is going to equal the following ratio. Where we have our concentration of our H plus protons which are on the product side. And we should raise this to an exponent of two because we have a coefficient of two in front of our H plus protons. So we raise that to a coefficient of two. And then in our denominator we want the concentration of our reactant ion which in this case is our concentration of our copper ion. And so we would have our concentration of our copper ion. And looking at the coefficient. When we multiplied our reaction by two, we have two moles of this copper ion. So we will raise it to an exponent of two. Now we also want to recognize that we have a second product that isn't an ion but rather is our sorry, second reactant, which is the hydrogen gas. And so we would multiply in our denominator by our pressure of hydrogen gas since it's our other reactant. And so we would say that this reaction quotient is equal to we have in our numerator our concentration of H plus which according to the prompt has a concentration of one molar. So we would plug in 1.0 Moller which is going to be squared. And then in our denominator we have our concentration of copper ions which we need to solve for here. And so we would say that that is going to be X squared times are pressure of hydrogen gas which according to the prompt we have a pressure of 1 80 M. So this would be times 1 80 M. And so continuing below. We would say that our reaction quotient is equal to one over X squared. And to simplify this further we would bring our X square to the numerator which would be the inverse. And so that would just be X to the negative second power. And so therefore Q. Is equal to X to the negative second power. And now that we have our value for the reaction quotient to solve for our concentration of copper caddy on. We can go ahead and now recall our nursed equation. So we'll do that below in the color black. So our equation we should recall is taking our self potential and setting that equal to our standard cell potential E degree. So subtracted from our new ernst equation constant which is a value of negative one times 0.592 divided by our electrons transferred N. And this is then multiplied by the log of our reaction quotient. Q. Now we want to go ahead and reorganize this to solve for the log of Q. And what that would give us is our self potential S l minus the standard cell potential. Not sell. And that is going to be divided by negative 0.0592, divided by our electrons transferred N. So we can place this in parentheses. So plugging in what we know we can say. And rather let's just write a equal sign here. We can say that this is equal to in our numerator. Our cell potential according to the prompt is 0.34 volts. So we have 0.34 volts subtracted from our standard cell potential, which above we stated is a value of 0.52 volts. So we plug that down below. And then in our denominator we have the equation constant negative 0.0592 divided by our electrons transferred. Where we said that that is equal to two. So we would say divided by two. And so in our calculators, we're going to plug in our quotient here and we would get a value equal to 6.0811. So just to simplify what we have, we have right now, the log of our reaction quotient is equal to 6.0811. And so we would get rid of that log term by taking Or rather by making both sides exponents. So we would say 10 to the log of Q is equal to 10 to the 6.811 power. This would allow us to cancel out that log term. And now we can say that Q. Is equal to a value of 1.2052 times 10 to the sixth power. And now that we have our value for Q. We can plug it into our reaction quotient where we said Q is equal to X squared to solve for our concentration of copper caddy on. And so we would say that therefore the concentration of copper Catalan is equal to. Or rather we can say can be found by taking cue Which we now know is just 1.20 52 times 10 to the six power. And setting that equal to X to the negative second power. And so to simplify this and sulfur X. We would take X and set that equal to the square root of 1/1 0.2052 times 10 to the six power. And so this is going to give us our concentration of copper or X value equal to a value of 9.1089 Times 10 to the negative 4th power. And so for our final answer, our concentration of copper Canyon is equal to 9.1089 times 10 to the negative fourth power. So this here would be our final answer to complete this example. So I hope that everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.

Ch.20 - Electrochemistry

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Ch.20 - Electrochemistry

Problem 112

Chapter 20, Problem 112

The cell potential of this electrochemical cell depends on the gold concentration in the cathode half-cell. Pt(s) | H₂(g, 1.0 atm) | H⁺(aq, 1.0 M) || Au³⁺(aq, ? M) | Au(s) What is the concentration of Au³⁺ in the solution if E_cell is 1.22 V?

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Identify the half-reactions for the electrochemical cell. The anode half-reaction involves hydrogen: \( \text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^- \). The cathode half-reaction involves gold: \( \text{Au}^{3+}(aq) + 3e^- \rightarrow \text{Au}(s) \).

Write the Nernst equation for the cell potential: \( E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \), where \( Q \) is the reaction quotient.

Determine the standard cell potential \( E^\circ_{cell} \) using standard reduction potentials: \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \). Use standard reduction potentials for \( \text{Au}^{3+}/\text{Au} \) and \( \text{H}^+/\text{H}_2 \).

Calculate the reaction quotient \( Q \) using the Nernst equation: \( Q = \frac{1}{[\text{Au}^{3+}]} \) since the concentration of \( \text{H}^+ \) is 1.0 M and \( \text{H}_2 \) is at 1.0 atm.

Rearrange the Nernst equation to solve for \( [\text{Au}^{3+}] \): \( [\text{Au}^{3+}] = \exp\left(\frac{nF(E^\circ_{cell} - E_{cell})}{RT}\right) \). Substitute known values to find the concentration of \( \text{Au}^{3+} \).

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Key Concepts

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

Nernst Equation

The Nernst Equation relates the cell potential (E) of an electrochemical cell to the concentrations of the reactants and products. It is expressed as E = E° - (RT/nF) ln(Q), where E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient. This equation allows us to calculate the cell potential under non-standard conditions, which is essential for determining the concentration of Au3+ in this scenario.

Standard Electrode Potentials

Standard electrode potentials (E°) are measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C) and provide a reference for predicting the direction of redox reactions. Each half-reaction has a specific E° value, which can be used to calculate the overall cell potential. In this question, knowing the standard reduction potential for the Au3+/Au couple is crucial for determining the concentration of Au3+ when the cell potential is given.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure of the relative concentrations of products and reactants at any point in a reaction. It is calculated using the formula Q = [products]/[reactants], where the concentrations are raised to the power of their coefficients in the balanced equation. In this electrochemical cell, Q will include the concentration of Au3+ and the concentration of H+ ions, and it is essential for applying the Nernst Equation to find the unknown concentration of Au3+ when the cell potential is specified.