A voltaic cell is constructed that is based on the following reaction: Sn2+1aq2 + Pb1s2 ¡ Sn1s2 + Pb2+1aq2 (a) If the concentration of Sn2+ in the cathode half-cell is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb2+ in the anode half-cell?

Question

Accepted Answer

welcome back everyone. The overall reaction for a voltaic cell is cobalt two plus plus nickel solid plus cobalt solid plus nickel caddy on the cell potential is 20.16 volts. When the concentration of cobalt Catalan in the cathode is 1.5 moller calculate the concentration of nickel cadmium in the an ode half cell for this case and were given our two half reactions below. So we should recognize that based on our overall reaction here we are given the correct order of our second half reaction because we begin with cobalt Catalan on the reacting side and we end with solid cobalt on the product side. However, in our overall reaction are nickel solids on the reacting side. We're in the half reaction given in the prompt, it's on the product side. So we're going to rewrite this first equation so that we now have solid nickel forming our products where we have our nickel canyon releasing two electrons as our second product. And we will still maintain that self potential of negative 0.26 volts. Now recognizing from the prompt, our concentration of our cobalt caddy on product or rather reacted here is under nonstandard conditions. We know that this reaction is occurring under nonstandard conditions because we want to recall that standard conditions which correspond to not sell and sorry, that's the zero. So is where we have at 25 degrees Celsius, a concentration of one molar a ph equal to seven and a pressure equal to 1 80 M. And so because our concentration is over one moller, we know that we are under nonstandard conditions. Next we want to recognize our noticed equation where we have our cell potential under nonstandard conditions equal to our self potential under standard conditions which we need to solve for which is subtracted from The quotient of 0.0592V divided by our electrons transferred end, which we will solve for, multiplied by the log of our reaction quotient Q. Where we would recall that Q is our concentration of products over reactant and only includes our Aquarius substances. So let's go ahead and figure out which of these half reactions occurs at the node and the cathode. So we can see that when we rewrote our first half reaction we now have electrons on the product side. Whenever we have electrons on the product side, this tells us that this reaction occurs as an oxidation which will occur at the anodes of our voltaic cell, meaning that we cannot label this first half reaction where we have electrons on the react inside as a reduction which will occur at the cathode of our voltaic cell which is also just given to us in the prompt because it tells us that cobalt in the cathode. So now that we have this labeled out, let's go ahead and add these two overall reactions because we can recognize that we have our electrons balanced. So we can just add these reactions together. So we will add this reaction here to cobalt two plus a Q gaining two electrons to form solid cobalt on the product side. And so we can cancel out our two electrons. We have it on the reactive side here and on the product side here. And what we're left with is going to be our overall reaction. Where we have solid nickel plus cobalt kati on producing solid cobalt plus our nickel Catalon. And we would say that our and this is we would say that our electrons transfer is equal to two N equals two. Now we need to figure out our standard cell potential for this overall reaction. So we want to recall that to calculate n not we're gonna take our cell potential of our cathode. Subtracted from the cell potential of our an ode. And what we're going to get when we plug things in. Because we're given our cell potential which we determined for our cathode is negative 0.28 volts. We're gonna plug that in as negative 0.28 volts subtracted from our cell potential of our note given in the prompt as negative 0.26 volts. So negative 0.26 volts. And from this difference we get our standard cell potential for our overall reaction equal to -0.02V. Now that we have our value for our cell potential under standard conditions, we're not gonna plug in everything that we know To find our concentration of our nickel catalon into our equation. So going into our equation, we have our cell potential under nonstandard conditions given in the prompt as 0.16V. This is set equal to our self potential under standard conditions which we just solved for as negative 0.02V which is then subtracted from The quotient of 0.0592V divided by our electrons transferred which we determined to be two, which is then multiplied by the log of our reaction quotient where again we only focus on a concentration of products divided by reactant and we're only using acquia substances. So we would have our concentration of our nickel Catalan which is what we need to find, divided by our concentration of our only Aquarius reactant which is our cobalt catalon. Now we can actually plug in our concentration of our cobalt cat and will do so in the next line because we're given that in the prompt. So what we'll have is .16V for our self potential under nonstandard conditions which was given to us equal to negative 0.02V subtracted from 0.0592V divided by two, multiplied by the log of our concentration of our Nickel Catalan which is what we need to sell four divided by our concentration of our cobalt Catalan equal to 1.5 molar according to the prompt. And so simplifying this will add our volt term to the to both sides. So we'll add 0.02V to both sides which will simplify the next line too. .18V equal to We have 0.0592V divided by two, multiplied by the log of our reaction quotient two plus And sorry, let's keep that in the bracket. So nickel two plus catalon divided by concentration of 1.5 molar. So simplifying this further, we will now have 0.18 volts equal to 0.0 to 96 volts. When we do that division, they're multiplied by the log of our reaction quotient are nickel Catalan concentration to our cobalt Catalan concentration of 1.5 molar. And this is still going to have that subtraction symbol in front of it coming from our step here, which it still should have. So this also needs a subtraction symbol. So we're going to continue by dividing both sides by negative 0.0-96V 0.02 96V. This will cancel out on the right hand side. And because we're dividing our units of volts our units of volts will also cancel out. And from that quotient we will have a value of negative 6.811 equal to the log of our reaction quotient. So nickel two plus concentration over 1.5 molar. And now to simplify both sides. We need to cancel that log term. So we're going to make both sides and exponents to the base of 10 which will get rid of that log term. And now we'll have in our next line of value of 8.297 times 10 to the -7 power equal to our concentration of nickel caddy on to our concentration of cobalt caddy on being 1.5 molar. And now recall that when we have diagonals in algebra we can just switch places places. So we will sorry, we'll switch places with the nickel and this term here on the left. So what we're going to have is that our concentration of our nickel caddy on Is equal to 8.297 times 10 to the negative 7th power over 1.5 Molar and sorry correction. So instead of What we did in the previous step, we're going to correct this by just multiplying both sides by 1. molar so that it cancels out on the right hand side. And we will finally get our concentration of our nickel caddy on equal to 1.2 times 10 to the negative six power moller. And this will be our final answer to complete this example as our concentration of nickel Canadian in the anodes half cell for the reaction. I hope everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

A voltaic cell is constructed that is based on the following reaction: Sn2+1aq2 + Pb1s2 ¡ Sn1s2 + Pb2+1aq2 (a) If the concentration of Sn2+ in the cathode half-cell is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb2+ in the anode half-cell?

Verified Solution

Key Concepts

Electrochemical Cells

Nernst Equation

Standard Electrode Potentials