The Nernst equation applies to both cell reactions and half-reactions. For the conditions specified, calculate the potential for the following half-reactions at 25 °C. (b)

Video transcript

welcome back everyone. The Nerdist equation is applicable not just to sell reactions but also two half reactions. Consider the following half reaction where two moles of mercury to reacts with two electrons to produce two moles of mercury plus carry on corresponding to a reduction potential of 120.92 volts. Using the nerds equation. What is its potential at 25 degrees Celsius. If mercury too, concentration is equal to mercury 12 plus concentration, Which all equals .115 Molar. So we're going to need to recall our nerds equation and recall that for our nerds equation, we have our non standard cell potential which is just represented as S. L. Equal to our standard cell potential as E not sell, Which is subtracted from the following quotient where we have .0592V divided by N. Which represents electrons transferred. And this is all multiplied by the log of Q. Which is our reaction reaction quotient. And recall that Q. Our reaction quotient is representative of our concentration of products over our concentration of react ints. So let's go ahead and get into this equation by plugging in what we know, we can say that our self potential under nonstandard conditions, Which we know is what we're solving for. Since we don't have concentrations of one molar, we have concentrations of .15 molar. So we're setting this equal to our self potential under standard conditions, which is given in the prompt as .92V and subtracted from 0. volts divided by N. R electrons transferred. Which we can see in the given half reaction is two electrons. So N would be equal to two and we would plug that in below and then we have the log of our reaction quotient where we would have the following quotient in our numerator we have our concentration of our product which in this case is our mercury one, so that's H G sub 22 plus Catalan. And we can see that in the equation. It has a coefficient of two. So we would recall that in our reaction, quotient coefficients from our equation will become exponents. So we raised this to a power of two. And now for our concentration of our reactant, which we see is mercury too. So that's HG two plus where we see we have a coefficient of two as well. So this would be raised to a power of two. And so plugging in the given concentration for Mercury two and Mercury one. In our next line we would have that are self potential under nonstandard conditions is equal to 0.92V subtracted from our quotient, 0.0592V over two, Multiplied by the log of our concentration of Mercury one, which is also .15 molar raised to the second power and then divided by our concentration of mercury two, which is 1.5 molar raised to a power of two. So simplifying this, we're going to ultimately be able to say in our next line that our standard or sorry, cell potential under nonstandard conditions is equal to 0.92 volts subtracted from 0.592 volts divided by two. Where because our numerator and denominator are the same in the log of Q. We can just say that we have the log of one Where next we would be able to simplify this. Quotient, multiplied by our log of one. Where in our next line we would find that this equals zero so that our self potential under nonstandard conditions is actually just equal to .92V. And so this would actually be our final answer. To complete this example. As our cell potential under nonstandard conditions when our concentration of mercury to and mercury one both equal 10.15 moller at 25 degrees Celsius. This is going to correspond to choice. Be in the multiple choice. I hope everything I argued was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.