The silver oxide–zinc battery used in watches delivers a voltage of 1.60 V. Calculate the free-energy change (in kilo-joules) for the cell reaction

Video transcript

Welcome back. Everyone used in handheld devices such as smartphones, lithium ion batteries can deliver a potential of 3.70 volts. Calculate the change in gibbs free energy for the reaction of this cell. Now we are given a cell where lithium reacts with Cobalt four oxide to produce lithium Cobalt three oxide. So if we analyze this reaction, we notice that lithium has an initial charge of zero, right? Because it's lithium solid, what about Cobalt or oxide? We have two oxygens, they each have a two negative charge which leaves us with a positive four charge on Cobalt. Now, for the ionic compound formed, we know that lithium would have a positive one charge because it belongs to the group one A, we have two oxygens with two negative charge and this leaves us with a positive three charge for Cobalt. So what we notice is that lithium becomes lithium plus, when we balance this equation, we get one electron. And if we just break it down, in simplest terms, we notice that Cobalt positive four, if we just think about the oxidation state rather than a charge would produce Cobalt three plus, let's make it ionic instead. So let's make it consistent with positive or positive, right? And we're adding one electron so the electrons match, right? We have a transfer. We want to recall that the change in the gibbs free energy is equal to the negative number of negative number of moles of electrons transferred multiplied by the parade constant. And the self potential. In this case, we have one mole of electrons paradise constant would be 96485 cool arms per mole of electrons. And we're multiplying by 3.70 volts, right? Because this is the cell potential. And when we calculate the result, we get negative 357 kilojoules, when we essentially divide the answer by 1000 to get kilojoules instead of joules, that's our final answer. Thank you for watching.