A concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.10 V at 25°C. What is the ratio of the Sn2+ concentrations in the two half-cells?

Video transcript

hey everyone in this example at 25 degrees Celsius. We're told that to zinc, zinc two plus have cells comprise the concentration cell with a cell potential of 20.32 volts. We determine the proportion of zinc two plus in the two half cells. So we should recall that the standard cell potential for a concentration cell E degree cell is equal to a value of zero volts. So what were given in the prompt is just our cell potential value Sl which is 0.32 volts. We next want to recall our nursed equation which is the cell potential S. L. Is equal to the standard cell potential degree cell which is then subtracted from our first equation constant 0.592 volts divided by N. Which would represent our electrons transferred, multiplied by the log of Q. Now it's important to write out how we produce our solid zinc here. So we take our zinc two plus cast iron And it gains two electrons to form solid zinc. We should recall that in our next equation Q. Is equal to the ratio between the ions of our half cells here. So what this means is that we would have in our numerator, the concentration of our zinc two plus an ode And then the concentration of zinc two plus of our cathode. In the denominator recall that reduction occurs that are an ode and oxidation or sorry, reduction occurs at our cathode and oxidation occurs that are an ode. So because we need to determine the value for this ratio for our final answer, we need to find the value for Q. So plugging in what we know for an ernst equation we're going to sulfur Q. Which would be the log of Q. Is equal to the difference between our self potential. E cell minus our standard cell potentially degree cell which is then going to be divided by -1 times 0.0592V divided by N. So this means we can say that the log of Q is equal to our cell potential given in the prompt as 0.32 volts subtracted from our standard cell potential for a concentration cell which we determined is zero volts. And then this is going to be divided by our nurse equation constant negative 0.592 volts. Which is also divided by N. Where N represents our electrons transferred. And we can see that we have two electrons that are transferred. So we would divide by two here. So we would further simplify and say that the log of Q is equal to a value of negative 10.8108. And now we want to cancel out that log term. So we're going to take the or make the right hand side and explain it to 10. So it can cancel out the long term. And this would allow us to say that Q is equal to a value of 1.5459 times 10 to the negative 11th power. So this would be the proportion of zinc two plus in our two half cells as our final answer. 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.