What is the Fe2+: Sn2+ concentration ratio in the following cell at 25 °C if the measured cell potential is 0.35 V?

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Welcome back everyone. We need to consider the following cell notation. We need to also determine the ratio between magnesium two plus Catalan and copper two plus carry on as a concentration ratio at 25 degrees Celsius is the cell potential is measured to be 2.74 volts. Were given the standard reduction potential for magnesium two plus carry on equal to negative 2.37 volts. And the standard reduction potential for copper two plus Equal to .34V. So let's begin by recalling that. For cell notation we have our physical barrier which separates both sides of our electro voltaic cell. And so we'll say B here is our physical barrier where we have first side A our first electrode which will be represented as a single line here. And this would be representative of our anodes where oxidation occurs. Whereas for C we have the second electrode which represents our cathode, where we should recall reduction occurs with this outlined, let's get into writing out our half reaction for each of our electrodes. So for the first half reaction we have according to the given cell, solid magnesium which becomes magnesium carry on. So we have MGs which forms Magnesium two plus and the release of two electrons. Now this edition of two electrons cancels out this net charge of plus two, meaning we have and that charge of zero so neutral on both sides of the reaction. And so we can confirm that are magnesium atoms are not only balanced, but the net charge of both sides of our reaction is balanced. Now let's move on to our second half reaction, which according to our given cell is for copper cat down forming solid copper. And so we would have see you This is a two plus Kati on which forms solid copper. Now we have a neutral charge on our product side. So we're going to add two electrons to our reactive side to cancel out that net plus two charge from our copper caddy on. And now we again have atoms and net charge balanced on both sides of the reaction. Now recognize that our oxidation state of magnesium in its standard form as a solid, is going to equal zero, whereas on the product side, magnesium has an oxidation state equal to its ion charge as plus two. And because the oxidation state increases, we would therefore label this first half reaction as an oxidation. Whereas our second half reaction, we see that our oxidation state of copper now on the reaction side is equal to its ion charge as plus two were on the product side, we have copper in its standard state as a solid, meaning it has an oxidation state of zero. We have a decrease in oxidation state, meaning that this reaction occurs as a reduction And so we can understand that our first half reaction is occurring at our anodes and our second half reaction occurs at the cathode electrode. Now we have that outlined. We want to because we have the same amount of electrons in both of our half reactions. Now add up these two half reactions to come up with an overall reaction. And so what we would have is we can cancel out our two electrons from both reactions. And what we're left with is solid magnesium Plus one mole of copper two plus catalon Forms one Mole of Magnesium two Plus Catalon. This is sorry a queue here. So we have MG two plus A. Q. And parenthesis. And our one mole of solid copper. Note that we are given from the prompt. Our standard reduction potentials for magnesium and copper caddy ons equal to our underlined values negative 2.37V and .34V. But we need to figure out our standard reduction potential E degrees cell for our redox reaction. Or we don't really even need to label it that way. But just for the entire electoral voltaic cell and recall that we would calculate that by taking the cell potential. The standard cell potential of our cathode subtracted from the standard cell potential of our a node which we can use the values given from the prompt. And so we would say that S. L. Is equal to our cell potential of our cathode given as the reduction of copper being 0.34 volts. So 0.34 volts subtracted from the cell potential of our anodes which we determined was our oxidation for the Given reduction potential of Magnesium two plus as point to point or sorry, negative 2.37V. So we have minus a negative 2.37V. This difference results in our cell potential for the electoral voltaic cell Under standard conditions equal to 2.71V. So now that we have this accompanying cell potential for our overall reaction we need to go ahead and determine the ratio between magnesium carry on and copper caddy on as a concentration. So we want to recall our nursed equation, recall that we would take the cell potential under nonstandard conditions, equal to our cell potential under standard conditions. E not so Which is subtracted from 0.0592V divided by N where N is our electrons transferred in both of our half reactions. And this is multiplied by the log of our reaction. Quotient Q where Q would represent our concentration of products in our overall reaction, divided by our concentration of reactant. Now, just to clarify something up here, I would like to clarify that when we found our value 2.71V. This is for the self potential under nonstandard conditions. So this is e cell here. Just to be clear actually. Um No, that was incorrect. So this is correct. As our self potential under standard conditions. And we know that because this is the general formula that you'll always follow the cell potential of your cathode minus the cell potential of your node under standard conditions. And going back to our prompts, we're told that we need to solve for our concentration ratio of our cat ions of magnesium and copper, meaning we don't know their concentrations. And so if this were under standard conditions we would assume that we have a concentration of one molar. But because we need to figure out the concentration, That means that this 2.74V represents our cell potential under nonstandard conditions. So E cell non standard. Sorry, let's write that where it's visible. So this value here is under nonstandard conditions. And so going back to our nearest equation, we can plug in for sl 2.74V as given from the prompt set this equal to S. L. So our cell potential under standard conditions which we which we calculated above as 2.71 volts. This is subtracted from 0.592 volts divided by our electrons transferred N. So for N we go back to our half reactions and we see that we canceled out two electrons from both of our half reactions. And so we would say N Is equal to two. So our denominator is two here and then this is multiplied by the log of our reaction quotient which in this case is based on our overall reaction. The concentration of our products being magnesium two plus Catalan which is our only Aquarius product and that would be the only product we include in the reaction quotient Q. And then in our denominator we have our concentration of our only acquis reactant being our concentration of our copper caddy on. So we're solving for that ratio of our concentrations of our ions. And so to simplify in our next line, We can begin by subtracting 2.71V from both sides. This cancels that out on the right hand side and that difference would be a value of 0.03 Which is equal to the right hand side of our equation where we would have negative and then our parentheses when we take the result of this quotient here, 0.0592V divided by two. We get a value of 0.0- which is still here, multiplied by the log of our ratio between magnesium catalon and copper cat ion concentrations. So now to simplify this further and isolate for the log of our ion concentrations, we're going to divide both sides by negative 0.296. So that it cancels out on the right hand side and that our left hand side will simplify to the result of the quotient being and sorry, this should be a negative sign here. So we get negative one point oh 13 51 which is equal to the log of our ratio between our concentration of magnesium caddy on to our concentration of our product copper caddy on or sorry our reactant copper caddy on. So we need to cancel out that log term. And so we can recall that we cancel out log by making it an exponents to the base of 10. And so for the left hand side we would make that also an exponent to the base of 10. So this cancels out log and what we would have is that our concentration of magnesium two plus two. Our concentration of copper caddy on. So this is our concentration ratio of our ions is equal to a value of positive 0. seven. And this would be our final answer. Now we need to consider sick fix. And based on our prompt we can confirm that we have a minimum of three sig fix. So we would round our final answer To also three cig fix as 0. 969. And so this would be our final answer for our concentration ratio between magnesium and our copper caddy on. This corresponds to choice be in the multiple choice. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

What is the Fe2+: Sn2+ concentration ratio in the following cell at 25 °C if the measured cell potential is 0.35 V?

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Key Concepts

Nernst Equation

Cell Potential

Reaction Quotient (Q)