For the voltaic cell, Sn(s)|Sn^2+(0.075 M) Pb^2+(0.600 M)|...

For the voltaic cell, $\operatorname{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.075 \mathrm{M}) \| \mathrm{Pb}^{2+}(0.600 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$ (a) what is $E_{\text {cell initially? }}$ (b) If the cell is allowed to operate spontaneously, will $E_{\text {cell increase }}$, decrease, or remain constant with time? Explain. (c) What will be $E_{\text {cell }}$ when $\left[\mathrm{Pb}^{2+}\right]$ has fallen to $0.500 \mathrm{M} ?$ (d) What will be $\left[\mathrm{Sn}^{2+}\right]$ at the point at which $E_{\text {cell }}=0.020 \mathrm{V} ?$ (e) What are the ion concentrations when $E_{\text {cell }}=0 ?$

For the voltaic cell, $\operatorname{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.075 \mathrm{M}) \| \mathrm{Pb}^{2+}(0.600 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$
(a) what is $E_{\text {cell initially? }}$
(b) If the cell is allowed to operate spontaneously, will $E_{\text {cell increase }}$, decrease, or remain constant with time? Explain.
(c) What will be $E_{\text {cell }}$ when $\left[\mathrm{Pb}^{2+}\right]$ has fallen to $0.500 \mathrm{M} ?$
(d) What will be $\left[\mathrm{Sn}^{2+}\right]$ at the point at which $E_{\text {cell }}=0.020 \mathrm{V} ?$
(e) What are the ion concentrations when $E_{\text {cell }}=0 ?$

Key Concepts

Redox Reaction Spontaneity and Equilibrium

Understanding redox reaction spontaneity involves recognizing that a positive cell potential indicates a spontaneous process under given conditions. As the reaction proceeds, the concentrations of reactants and products change until equilibrium is reached, which can lower the cell potential or bring it to zero when no net reaction occurs.

Nernst Equation

The Nernst equation allows for the calculation of an electrode’s potential under non-standard conditions by incorporating the effect of ion concentration (or pressure) on the reaction quotient. It is central to understanding how the cell potential adjusts based on changes in the concentration of reactants or products, enabling the prediction and control of cell behavior during operation.

Concentration Effects on Cell Potential

The impact of ion concentrations on cell potential is a key concept in electrochemistry. Changes in the concentrations of the ionic species involved in the half-reactions affect the reaction quotient, and through the Nernst equation, directly influence the overall voltage produced by the cell. This principle explains why and how the cell potential adjusts over time as the reaction proceeds.

Voltaic Cells

Voltaic cells, also known as galvanic cells, convert chemical energy into electrical energy through spontaneous redox reactions. They consist of two half-cells connected by a salt bridge, and the cell potential is determined by the difference between the electrode potentials of the cathode and anode. This concept is foundational for understanding how electrochemical reactions generate current and work in batteries.

Standard Electrode Potentials

Standard electrode potentials refer to the inherent tendency of a chemical species to be reduced under standard conditions (1 M concentration, 1 atm pressure, and 25°C). These values are crucial because they serve as reference points for calculating cell potentials under non-standard conditions using the Nernst equation, and help predict the direction of redox reactions.

Transcript

00:01 Hello, so today we're going to be looking at this situation right here, where we have a voltage cell.

00:10 And we can see at the anode, we have tin being oxidized.

00:16 And then at the cathode, we have lead being reduced.

00:21 So basically, we could rewrite, we can write both of the half reactions here.

00:39 So tin is being oxidized and lead is being reduced.

00:53 And we have two moles of electrons moving around.

01:02 So how about we find out what the cell potential is initially for this reaction? and let's just write out the full reaction, the equation for this reaction that's taking place.

01:47 So at 25 degrees celsius, our cell potential is given by the standard cell potential minus 0 .0257 over z, which is the moles of electrons times the natural log of q.

02:27 So what is the standard cell potential? well, we know that, let's take a look at here.

02:39 So the cell potential would be the reduction potential of what's being reduced, in this case led, which is negative 0 .125, subtracting the reduction potential of what's being oxidized, in this case, that would be basically the reduction potential of a tin ion being reduced to solid tin, which is negative 0 .137.

03:21 And that would give us 0 .012 volts.

03:39 And so we know our standard cell potential, and we know that the moles of electrons is 2.

03:59 So what would be q? well, if we look at our equation right here, it would be the concentration of tin over the concentration of lead.

04:10 Our concentration of 10 is 0 .075, and our concentration of lead is 0 .6.

04:29 And so now if we solve this, we will get 0 .039 volts.

05:03 So that's with the react initial concentrations we've been given.

05:13 This would be our cell potential.

05:16 So what would happen if we allowed this process to continue? so if we let this process go, we would start to see a decrease in the concentration of the lead and then increase the concentration of the tin, and that would cause our q to increase, become less negative, and that would cause our cell potential to decrease.

05:53 So our cell potential would decrease with time.

06:08 So now let's say that the concentration of lead, has fallen to 0 .5 0 -0 malarity.

06:32 So if we look at this equation right here, we see, for every mole of lead that reacts, we form one mole of tin.

06:45 So if 0 .1 molar of lead has reacted, we've formed 0 .1 molar of 10.

06:55 In.

06:57 So what would our cell potential be at that point? well, we just have to plug in.

07:05 So our standard cell potential hasn't changed.

07:16 It's still 0 .012.

07:27 We still have two moles of electrons moving around.

07:31 But our queue has changed.

07:34 So instead of it being 0 .075 over 0 .6, it's going to be.

07:42 0 .175 over 0 .5.