A Cu/Cu2+ concentration cell has a voltage of 0.22 V at 25 °C. The concentration of Cu2+ in one of the half-cells is 1.5x10^-3 M. What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell is the lower of the two concentrations.)

Video transcript

All right. Hi, everyone. So this question says that a copper to copper two positive concentration cell has a voltage of 0.22 volts at 25 °C. The concentration of copper two positive in one of the half cells is 1.5 multiplied by 10 to the negative third molar. What is the concentration of copper two positive in the other half cell? Assume the concentration in the unknown cell is the lower of the two concentrations. And here we have four different answer choices labeled a through D proposing different molar quantities. So before we go ahead and proceed with any relevant calculations, let's first go ahead and write out the relevant cell reaction. In this case, it's going to involve the redox reaction involving elemental copper and copper two positive. So the first half reaction is when copper two positive gains to electrons, they yield elemental copper. And so the standard potential for this reaction is equal to positive 0.34 volts. Now this is the reduction half reaction because electrons are being gained. So the other half reaction is its oxidation or in other words, the opposite process were elemental copper loses two electrons to yield copper two positive. And so the standard potential for the second half cell. In other words, the oxidation process is negative 0.34 volts. So when I combine both of these half reactions to yield the total cell reaction, elemental copper can be eliminated as it's present on both sides of the equation or rather on both sides of the Arab. And the same thing can be said for the two electrons. So the final reaction becomes copper to positive during the reduction process yield copper two positive oxidized. And so the E standard for the complete reaction is going to be the sum of the standard potentials for both half reactions. And this happens to be zero volts now because there is a mention of the standard potential and the voltage of the concentration. So we are going to need the Nernst equation. Recall that the Nernst equation does involve the reaction quotient otherwise known as Q for this reaction, Q is equal to the concentration of the oxidized copper I I positive, divided by the concentration of reduced copper, two positive. And this is because the reaction quotient is equal to the concentration of products divided by the concentration of reactants. And so with this information in mind, we can go ahead and recall the Nernst equation. So recall that the nurse equation states that the potential of the cell or E is equal to the standard potential of the cell subtracted by the following. That's 0.0592 divided by N which is the number of electrons being transferred multiplied by the log of the reaction quotient. So in this case, we're going to have to rearrange the nernst equation so that the logarithm of Q can be isolated and solved for. So this means that the logarithm of Q is equal to the difference between the potential of the. So and the standard potential multiplied by negative one and then divided by 0.0592 divided by N. So at this point, we can go ahead and substitute for the values determined previously eight, the potential of the cell that was provided in the text of the question was 0.22 volts and the standard potential of the cell was zero volts. And so we take the difference between both quantities and multiply that by negative one, before dividing by 0.0592 divided by two, which represent the two electrons transferred during this process. After evaluating this expression, the logarithm of Q is equal to negative 7.4324. And so to solve re cue specifically, we're going to raise 10, let's 10 to the power of negative 7.4324 which is equal to 3.6946 multiplied by 10 to the power of negative eight for the reaction quotient. So now that we have a numerical value for the reaction quotient. We can use this to solve four the concentration of copper two positive in one of our half cells. So here once again, Q is equal to the concentration of oxidized copper I I positive divided by the concentration of reduced copper two positive. Now one quick note before we go ahead and solve this last part of the question, right, we are given the concentration of copper two positive in one of the half cells without necessarily specifying which one. But we do know that we have to assume that the concentration of the unknown is the lower of the two concentrations. So if we consider the expression for the reaction quotient the reaction quotient itself 3.6946 multiplied by 10 to the power of negative eight. This quote this quantity, right? This quotient is less than one. And because this quantity is less than one, the value of the numerator must be smaller than the value of the denominator. So what we're solving for is the lesser of the two quantities which would be the concentration of oxidized copper two positive. So this means that the concentration of oxidized copper I I positive is equal to Q multiplied by the concentration of reduced copper to positive. So let's go ahead and solve for that. So here Q is equal to 3.6946 multiplied by 10 to the power of negative eight. And this is multiplied by the concentration of reduced copper two positive, which was given to us in the beginning as 1.5 multiplied by 10 to the power of negative three. So after multiplying these quantities together, the concentration of oxidized copper two positive is equal to 5.5419 multiplied by 10 to the power of negative 11 molar. And after rounding to two significant figures, this yields 5.5 multiplied by 10 to the power of negative 11 molar. And there you have it. So scrolling up here to consider our answer choices are answer corresponds to option D in the multiple choice and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.