Write equations for the half-reactions that occur in the electrol- ysis of molten potassium bromide.

Video transcript

Hey everyone in this example. We need to determine for the electrolysis of pure molten potassium bromide to determine the products that could potentially form. So what we should recall is that for the electrolysis of a pure molten salt, our an ion should be oxidized and our cat ion should be reduced. So potassium bromide consists of the following ions where we have the K plus, kati on And the BR -5. So beginning with our carry on, we have K plus which is an ionic liquid which should form solid potassium. Looking at our ion or an ion, rather we have the B R minus an ion which is a liquid. An ion which will form the bro, mean di atomic molecules in liquid form. And looking at this second reaction here, we want to recognize that we have two atoms of bromine. So we're going to balance this out by placing a coefficient of two in front of our react inside here, which would now give us two atoms of bromine here. However, we want to recognize that we need a net neutral charge on both sides of our reaction. Or rather We would recognize that we have a coefficient of two in front of this -1 here. So this would give us a net charge on the reactive side of -2. So we need a net -2 charge also on the on the product side here. And so right now we have a net neutral charge because we have a neutral compound of liquid bromine. So we need to go ahead and make this Have a net charge of - by adding two electrons here. And this would make sense that we added two electrons to the product side because our an ion should occur as an oxidation. And we recall that when we add electrons to the product side, we have an oxidation occurring. Now in order to add these two reactions up, we need the amount of electrons to match both sides of our equation, Meaning we would need to multiply this first equation by value of two. So in doing so, we would form a new equation here and I'll just write it below where we would have two atoms of the potassium carry on in liquid form, producing two moles of our potassium solid. And so now we have a net charge of plus two on the reactant side where we have a net charge of zero on the product side. So we would go ahead and compensate for this change by creating a negative charge. Or sorry, a net charge of zero on the reacting side by losing to elect or adding two electrons here. So we would say plus two electrons. And this would now cancel out that Plus two charge here because we added two electrons so that now we have a net charge of zero on both sides of this equation. And because it is clear that we added two electrons here to our react inside, we would recall that when we add electrons to the reactive side, this reaction would occur as a reduction. So now that we have our atoms as well as our charges on both sides of our following reactions here balanced, we can go ahead and finally add these reactions together. And so in adding these two reactions together, we're going to form one equation where we have two moles of our potassium plus carry on in liquid form Plus two moles of our bromide, an ion in liquid form. We would go ahead and cancel out the electrons since this electron is on the product side. And these two electrons are on the reacting side and the other equation. And then we're going to continue to our products where we have two moles of our potassium solid Plus one mole of our bro, mean liquid. And so we can clearly see our products here which will be our two moles of our potassium solid and our one mole of our blooming liquid. So what's highlighted in yellow represents our final answers as the products that will form from the electrolysis of pure molten potassium bromide. I hope that everything I reviewed was clear. If you have any questions leave them down below. Otherwise, I'll see everyone in the next practice video