Make a sketch of the voltaic cell represented by the line nota- tion. Write the overall balanced equation for the reaction and calculate Ec°ell. Sn(s) | Sn2+(aq) || NO( g) | NO3-(aq), H+(aq) | Pt(s)

Video transcript

hey everyone in this example, we need to calculate the standard cell potential for the following cell. So we should recognize that. Based on this given cell, we have two half reactions. Our first half reaction being our solid iron forming the Iron three Plus Cat Ion. Whereas our second half reaction is going to be our chlorine gas which forms our chloride an ion. Now we want to recognize that we need the atoms to be balanced. Here are atoms of iron are balanced in the first half reaction. However, we need to add a coefficient of two in front of our chloride an ion so that we have two atoms of chlorine with two on the reactant side. Our next step is to make sure that our charges are balanced. So here we have a net charge of zero for our neutral atom of solid iron. However, on the product side we have a net charge of plus three. And so we need to cancel out this net charge of plus three. And in order to do so, we would add three electrons to the product side here. Now because we added electrons to the product side, this means that this half reaction occurs as an oxidation. Now moving on to our second half reaction, we have a net charge of zero for our react inside however, we have a net charge of the coefficient of two times negative one, which would be minus two on the product side. So to cancel out this net charge of minus two Or rather to get a net charge of -2 on the reactant side, we're going to go ahead and expand our reacting side so that we add two electrons here so that it has a net charge of -2. And so we would have a net charge of -2 on both sides, meaning our charges are balanced for the second half reaction. And because we recognize that we added electrons to the reactive side of this half reaction here. This half reaction is going to occur as a reduction. We should recall that our oxidation half reaction will occur at the anodes in our cell. And then for our reduction half reaction, we should recall that this occurs at the cathode of our cell. So we can label here that this is our cathode. We recall that this double barrier here. This is a barrier of our cell and then this portion of our cell is are an ode. So our next step is to refer to our standard so reduction potential table. So it's the standard reduction potential table actually. And when we refer to this table either online or in our textbooks, we're going to first find for the oxidation of solid iron. The self potential value Which is going to equal according to this table, a value of -0.036V for our node. And so doing the same to find our cell potential for the oxidation of chlorine gas. We would see that according to this table, we're going to have a voltage of 1.36V at our cathode. And so now we can go ahead and calculate our standard cell potential which is equal to the self potential of our cathode. Subtracted from the cell potential of our a note. And so we would plug in our values from above where we would have for our cathode. We said that's equal to value of negative 0.036V which is subtracted from our self potential of our anodes which above we stated is 1.36V. I'm sorry about that. I plugged that in backwards. So we have our cell potential of our cathode added to our cell potential of our anna to get our standard cell potential. And above we stated, our self potential of our cathode is equal to 1.36V Added to our self potential of our anodes. Which above we stated a 0.2.036V. So adding these two values up, we're going to get our self potential our standard cell potential rather equal to a value Of about 1.40V. And so this would be our final answer here for our standard cell potential of our given cell above. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below. Otherwise I will see everyone in the next practice video