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Galvanic cells and changes in free energy

Video transcript

we already know that for this galvanic cell solid zinc is oxidized to zinc two plus ions oxidation occurs at the anode so this electrode is our anode and copper two-plus ions are reduced to solid copper reduction occurs at the cathode so this electrode must be our cathode if we start with a one molar concentration of zinc sulfate that means in solution our initial concentration of zinc two-plus would be 1 molar and we have a 1 molar concentration of copper sulfate which means we're starting with a concentration of copper two plus of 1 molar so we're under standard conditions I have 1 molar for our concentrations we're at 25 degrees C we have solid we have solid pure electrodes and we know that the standard cell potential so the cell potential under standard conditions for this for this cell is positive 1 point 1 0 volts so we have a spontaneous redox reaction which produces a current so electrons flow in our wire and we get a voltage we have a standard cell potential in one of the videos in electrochemistry we took our standard cell potential and from it we calculated Delta g0 so we used this equation we said Delta g0 is equal to negative n F e0 so we plugged in our standard cell potential and we got Delta G 0 so Delta G 0 is the change in free energy under standard conditions and we got negative 212 kilojoules let's look at the equation for the change in free energy from thermodynamics and let's analyze our galvanic cell so the change in free energy is equal to the standard change in free energy Delta g0 plus RT natural log of Q so remember this Delta G is the instantaneous difference in free energy between your reactants and your products Delta g0 is the change in free energy under standard conditions r is the gas constant t is temperature in Kelvin and Q is your reaction quotient so let's think about what the reaction quotient is for this reaction so for this reaction our spontaneous redox reaction Q has the same form as the equilibrium constant concentration of products over concentration of reactants and you leave out pure solids so that would be the concentration of zinc two-plus right that's our product here we're leaving out solid copper over the concentration of copper two plus and leaving out solid zinc so over the concentration of copper two plus at this instant in time we plug in our concentrations right the concentration of zinc two plus is one molar so we have one point zero the concentration of copper two plus is also one molar so we're under standard one our standard conditions here so Q is equal to one at this instant and we plug in Q is equal to one into our equation and we see that the natural log of one right natural log of one is equal to zero so this this term goes to zero and under standard conditions the change in free energy delta-g right the change in free energy Delta G is equal to the change in free energy under standard conditions Delta G zero and that's equal to negative 212 kilojoules and if you wanted to write per mole of reaction alright you could do that right here and so this makes sense because we're under standard conditions right we're at one molar for our concentration so the change in free energy is equal to is equal to Delta G zero notice that Delta G is negative so we know this is a spontaneous reaction right this is a spontaneous reaction under standard conditions so current flows we get a voltage so let's go back up to here all right so we get a voltage we get a voltage at this moment in time so that the reaction goes to the right to make more of our products right we're going to make more of our products here what happens to Q as the reaction proceeds to the right well we're increasing the concentration of our products we're increasing the concentration of zinc two plus ions at the same time we're decreasing the concentration of copper two plus ions so Q increases as the reaction proceeds to the right so as we make more products right Q increases so we get an increase in Q what happens to Delta G what happens to the instantaneous difference in free energy between our reactants and our products let's go ahead and plug that in to our equation so let's make up a number let's say that Q increases from we started over here with 1 let's say that Q increases to 10,000 so or to pick a big number here so Q goes up to 10,000 so we have far more products than we do reactants what is Delta G Delta G is equal to Delta G 0 is the standard is the standard change in free energy so under standard conditions this is negative 212 kilojoules so that's negative 212 and this would be this would be negative 212 thousand joules if we convert kilojoules to joules plus plus we have R which is the gas constant so let me go back up to here R is the gas constant which is 8.314 joules per mole times Kelvin so you could make this joules per mole here so your units balance out moles per mole of reaction and then we have the temperature so the temperature of our reaction let's remind ourselves of that so we go back up to here we're at 25 degrees C so 25 degrees C is 298 Kelvin so this is 298 Kelvin so Kelvin would cancel out here and this is times the natural log of Q and now we've changed Q we've said ok this is pick a number that's pretty big so Q is 10,000 so you plug in the natural log of 10,000 so we have obviously a huge number of products compared to reactants at this moment in time so at this moment time what is Delta G what is the instantaneous change in free energy between our reactants and our products so we need to do that calculation here so let's find the natural log of 10,000 so we have that we're going to multiply that by 298 and we're going to multiply that by 8.314 and we're going to add that number we're going to add that number to negative 212 thousand so we get that Delta G is equal to negative one hundred and eighty nine point two let's say kilojoules so I'll make that into kilojoules so Delta G is equal to negative one hundred and eighty nine point two kilojoules per mole so I'm not really too concerned about the exact number I'm just trying to point out what happens as you change Q all right so as we increased Q as we went from Q is equal to 1 to Q is equal to 10,000 so what happened to Delta G we went from negative 212 kilojoules to negative 189 point to kilojoules so we're getting closer to zero we're getting closer to equilibrium but at this instant right at this instant right here when Q is equal to 10,000 we still have a negative value we're still have a negative Delta G I should say so the reaction is still spontaneous the reaction is still spontaneous we're still going to generate a current we're still going to make more of our products we're still going to have a voltage at this moment in time what would be the voltage at this moment in time what would be the instantaneous cell potential so the instantaneous cell potential is e we could find that using our equation that relates Delta G and E Delta G is equal to negative NF e so if we plugged in if we plug in this if we plug in Delta G into here all right we know that n is the number of moles of electrons that are transferred F is Faraday's constant and an E is our instantaneous cell potential to save time I won't do the calculation here but if you do that calculation you will get that your instantaneous cell potential is equal to 0.98 volts positive 0.98 volts so the reaction is spontaneous we're still producing a voltage we are still producing a voltage here so notice that a voltage has decreased a little bit right in the previous example under under standard conditions at that moment in time the voltage was 1.10 volts so we've lost a little bit of voltage here we're down to 0.98 but think about how large that number is right we have so many of our products compared to our reactants and we're still getting a pretty decent voltage so approximately 1 volt so pretty close to the original one point one volts what happens at equilibrium alright so what happens at equilibrium we know that at equilibrium the reaction quotient Q is equal to the equilibrium constant K and at 25 degrees C K is equal to one point five eight times 10 to the 37th so for this reaction at 25 degrees C this is our equilibrium constant let's plug in our equilibrium constant into our equation for Delta G to see what we get so Delta G is equal to Delta g0 which was negative 212 kilojoules per mole plus R which is 8.314 times the temperature we're still at 25 degrees C so this is 298 K and this time we're putting in the natural log of K so we're plugging in our equilibrium constant for Q so natural log of one point five eight times 10 to the 37th so let's do that math now we have the natural log of one point five eight times 10 to the 37th and we're going to multiply that by 298 298 + 8.314 and that gives us that gives us if we're out that that's that's positive 212 kilojoules per mole so positive 212 kilojoules per mole I'm not worried about the exact number here because I rounded this this is rounded as well the point is that Delta G is equal to zero at equilibrium we already know that we already know Delta G is equal to zero at equilibrium this shows you that we have negative 212 kilojoules per mole so about negative 212 kilojoules per mole and approximately 212 kilojoules per mole over here at equilibrium these will cancel out and give you Delta G is equal to zero so at equilibrium at equilibrium we know that Delta G is equal to zero there is no difference in free energy between your reactants and your products so let's think about the voltage of our cell at equilibrium well if Delta G is equal to zero we plug that into here and therefore the cell potential e must be equal to zero right if this is equal to zero then this is equal to zero so the cell potential is equal to zero volts the voltage is zero when our redox reaction comes to equilibrium and so therefore the cell dies your battery is dead so hopefully this helps you understand galvanic cells in terms of thinking about changes in free energy