A voltaic cell consists of a strip of cadmium metal in a solution of Cd1NO322 in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with Cl2 gas bubbled around the electrode. A salt bridge connects the two beakers. (c) Write the equation for the overall cell reaction.

Video transcript

Welcome back everyone. A galvanic cell is composed of two beakers connected by a salt bridge. One beaker is made up of a strip of nickel metal submerged in a nickel to nitrate solution. The other beaker is made up of a platinum electrodes submerged in a potassium flooring solution wherein our florian gas is bubbled around the electrode and give the overall cell reaction equation. So we're going to need to write out our redox reactions involved here. Beginning with our first where we have solid nickel metal which releases as a product our nickel to Pluskat ion. I'm sorry, that's nickel two plus carry on its acquis plus our two electrons as products for our next half reaction, we focus on nitrate which we call us N +03 minus one. We're on our product side, we form nitrogen monoxide gas. For our third half reaction, we focus on our potassium catalon, we're on our product side. We'll have solid potassium and for our fourth half reaction we focus on our flooring gas where on our product side we have our fluoride an ion. Now we need to make sure that these are balanced reactions. So we have our atoms for nickel balanced in the first half reaction as well as the charge balance because we have a net charge of zero on both sides. The plus two charges canceled out by the two electrons here for our second half reaction, we have three moles of oxygen on the reaction side and one mole of oxygen on the product side. Recall that we use water to balance oxygen and so we're going to add two moles of water to the product side here for the second half reaction. So now we have introduced formals of hydride on the product side and recall that we use hydride to balance hydrogen. And so we're going to expand the react inside here for the second half reaction and add four moles of r H plus catalon hydride. Now focusing on the electrons here, we have an overall charge of on the reactant side minus one plus four, which would leave us with minus three On the product side, we have a net charge of zero. So we need to cancel out this -3 or sorry, plus three overall charge here on the reactant side. So we're going to expand this again by adding a third reactant which will be plus three electrons. Moving on to our third half reaction, we have a net charge of plus one on the reactive side and a net charge of zero on the product side. So we're going to cancel out that net charge of plus one by doing the same thing and adding electrons to the reacting side. So we'll add one electron here. Now going to our fourth half reaction, we have two atoms of flooring on the reactive side and one on the product side. We're going to place a coefficient of two on our product side. Now bouncing out our flooring atoms but now we have a net charge of minus two on the product side here. So to bounce out that charge, we're going to get a net charge of minus two on the reactant side by also adding two electrons here on the reactant side. So now we have our charges and Adams bounced in each of our Redox reactions here. We now want to focus on our ion pairs in each of our beakers. So for bigger one we have our ion pair being our nickel metal and our nickel catalon. And for beaker too, we have our ion pair being our floor in gas and our two moles of our fluoride. An eye on here. Now focusing on bigger one. We see that we have electrons on the product side here, which we want to recall and let's use a different color. Let's use green. We have two electrons on the product side recall that when electrons are on the product side, This tells us that this reaction occurs as an oxidation, which we would recall occurs at the anodes of our voltaic cell. We see we have two electrons on the react inside here for the floor in half reaction, which tells us that this reaction occurs as a reduction where we recall. This means occurs. This reaction occurs at the cathode of our voltaic cell. And so, looking up in our textbooks, our cell potential for the oxidation of solid nickel, we get a standard cell potential equal to a value of -0.28V. And now for the standard cell potential of our reduction of flooring gas, we have a value in our textbooks of positive 2.87V. Now recognize also that in both of these half reactions we both have or they both have two electrons. So our electrons are balanced. And now we can add up these two half reactions to get one overall reaction. So what we'll have is we're adding up our solid one mole of solid nickel which produces one mole of our nickel cadmium Plus our two electrons to our second half reaction being our flooring gas Plus two electrons to form two moles of our fluoride an ion. And so we want to cancel out those repeat of our two electrons here on the product side and then here on the react inside and everything else that were left over with will be our overall reaction. Where we have one mole of solid nickel reacting with one mole of our flooring gas. We're on our product side. We have one mole of our nickel cat ion Plus our two moles of our fluoride an ion. And this would be our final answer as the overall cell reaction equation. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video