A voltaic cell consists of a Zn/Zn 2+ half-cell and a Ni/Ni 2+ half-cell at 25 °C. The initial concentrations of Ni 2+ and Zn 2+ are 1.50 M and 0.100 M, respectively. b. What is the cell potential when the concentration of Ni 2+ has fallen to 0.500 M?

Hey everyone in this example, we need to calculate the cell potential when the silver plus one caddy on has already dropped to a concentration of 10.50 Moeller. So we're going to first write out our two half reactions are first reaction will show our formation of solid silver where we have the silver plus one cat ion, Gaining an electron to form one mole of solid silver. And then for our second half reaction, we're going to have solid magnesium, which breaks down to the silver two plus caddy on which forms solid magnesium by gaining two electrons. Now, when we look up in our standard electoral potential table for the reduction of solid silver, we would see that we have a self potential value equal to .80V. And when we look up the cell potential reduction table for the oxidation of solid magnesium, we have a self potential value equal to negative 2.37V. We should recognize that this negative value is referring to this half reaction occurring at our node as an oxidation. And this positive value means that this half reaction occurs at our cathode as a reduction. Now we should recognize that the electrons do not match. We have one electron in the first half reaction and two in the second half reaction. So we're gonna take this first half reaction and multiply it by two. And what we would get is two moles of the silver plus one cat, I am added to two electrons produces two moles of the solid silver as a product and now we're going to be able to add this to our second equation there And this says two plus. So adding these two equations together, we're gonna end up with just one full equation where we have first, we're going to just cancel out these two electrons because we have it here on the reactant side and then here on the product side. So what we would get as a single reaction is two moles of the silver plus one cat ion plus one mole of solid magnesium produces as a product. We have two moles of solid silver plus one mole of the magnesium two plus catalon. Our next step is to calculate our standard cell potential for our reaction. And we should recall that that is calculated by taking the cell potential of our cathode subtracted from the cell potential of our an ode. So what we would get is for our cathode, we determined that we have a cell potential of 0.80V subtracted from our cell potential of Iran, in which we determined is negative 2.37 volts. And this difference gives us our standard cell potential equal to a value of 3. volts. Now, based on our reaction, we should recognize that one mole of magnesium uses two moles of the silver plus one Catalan, which should increase the polarity of the magnesium two plus carry on by half of our decrease of our silver plus one, carry on. So we want to find our polarity of the magnesium two plus catty on here. So to calculate that we would take the polarity of our magnesium magnesium two plus half cell. Given in the prompt as 0.0 to 50 moller. And we're going to add this to the polarity of our silver silver plus one Catalan half cell, which according to the prompt is 1.32 moller. And according to the prompt, the polarity of silver plus one cat and will drop 2.50 moller. So we're going to subtract this value by 0.50 moller and then we're going to divide this by two. So what this would give us is our polarity of magnesium two plus equal to 0.4 35 moller. Next we want to look at our reaction quotient Q. Where we would look at the concentration of the ratio or the ratio of the concentration between each ion in our half cell. So first the magnesium two plus ion which occurs at our an ode half cell and then divided by the concentration of our silver plus one cat, which occurs at our cathode half cell. And we're going to raise this to an exponent of two here because it has a coefficient of two in our equation that we outlined above. So this occurs again at our cathode or oxidation half reaction. So plugging in the concentrations from above. We would get for our concentration of magnesium two plus we set above that, that is 0.4 35 molar. And then in our denominator according to the prompt, the concentration of silver plus one as a Catalan is 10. moller. And this should be squared because we have that coefficient of two. So this would give us our reaction quotient equal to a value of 1.74. And now we can use this value to calculate our cell potential by utilizing our nursed equation which we should recall is taking our cell potential equal to our standard cell potential subtracted from our first equation constant 0.0592V divided by our electrons transferred n multiplied by the log of our value for Q. So plugging in what we know we're going to solve for SLR cell potential which is equal to our standard cell potential. Which above we calculated here as 3.17V. This came from our values in our textbook and then this is subtracted from again our quotient where we take the first equation constant 0.0592V divided by our electrons transferred. And above we can see that we only transferred two electrons. So that is for our reduction half reaction will plug into in the denominator here. And then this is multiplied by our log of Q. Which we said Q is equal to a value of 1.74. And so simplifying for our self potential, we would get a value equal to 3.16V. And so this would be our final answer here. To complete this example as our self potential when the concentration of the silver plus one carry on drops 2.50 moller. 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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Tro 5th Edition

Ch.20 - Electrochemistry

Problem 77b

Chapter 20, Problem 77b

A voltaic cell consists of a Zn/Zn²⁺ half-cell and a Ni/Ni²⁺ half-cell at 25 °C. The initial concentrations of Ni²⁺ and Zn²⁺ are 1.50 M and 0.100 M, respectively. b. What is the cell potential when the concentration of Ni²⁺ has fallen to 0.500 M?

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Identify the half-reactions for the voltaic cell: Zn -> Zn^{2+} + 2e^- and Ni^{2+} + 2e^- -> Ni.

Write the Nernst equation for the cell potential: E = E^0 - \frac{RT}{nF} \ln Q, where Q is the reaction quotient.

Calculate the standard cell potential (E^0) using standard reduction potentials: E^0_{cell} = E^0_{cathode} - E^0_{anode}.

Determine the reaction quotient (Q) using the concentrations: Q = \frac{[Zn^{2+}]}{[Ni^{2+}]}.

Substitute the values into the Nernst equation to find the cell potential at the given concentrations.

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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 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.

Cell Potential

Cell potential, or electromotive force (EMF), is the measure of the energy per unit charge available from a voltaic cell. It indicates the tendency of the cell to drive an electric current through an external circuit. The potential is influenced by the concentrations of the reactants and products, and it can change as the reaction proceeds, which is crucial for understanding how the cell operates over time.

Half-Cell Reactions

In a voltaic cell, half-cell reactions occur at the anode and cathode, where oxidation and reduction take place, respectively. For the Zn|Zn2+ half-cell, zinc is oxidized to Zn2+, while in the Ni|Ni2+ half-cell, Ni2+ is reduced to nickel. Understanding these half-cell reactions is essential for calculating the overall cell potential and determining how changes in concentration affect the cell's performance.