 So, we have this question here which is asking for three things. First to construct a galvanic cell based on this information to write the cell reaction and to calculate the cell potential. And the information that is given are these two half reactions and their standard reduction potentials. So, you can pause this and give this a try and we'll continue in a few seconds. Okay. So, the first thing is by looking at this information, we want to identify what is going to get oxidized and what is going to get reduced. So, for that we use these standard reduction potentials and we know that higher this value, the more easy it is for reduction to take place. So, in this case for this ion with a plus 3 charge, the reduction potential is higher which means this will be easily reduced. So, we can write this half reaction like this where we see that this Fe3 plus accepts an electron to go to Fe2 plus. Now, for the oxidation reaction, we write the reverse of this reaction with zinc which we can write like this. So, these were the half reactions. Now, to get the cell reaction, we need to add these half reactions. But before that, because there is one electron on the left hand side and two of them on the right hand side, we need to multiply this reaction by 2. So, we get something like this and now if we add both of these half reactions, we get this reaction which is going to be our cell reaction. Now, another part is asking us to calculate the cell potential. So, since we have the standard reduction potentials of both of these reactions, I've written them down here. So, we were given the reduction potential for zinc. But here, we are flipping this reaction and we are writing it as oxidation reaction for our cell. So, the potential now becomes positive 0.76 volt. And now to get the cell potential, all that we do is add both of these. So, we get the value to be 1.53 volts. Now, one common mistake here is to treat these potential values like enthalpy. So, since we multiplied this by 2 to get the cell reaction, if you also multiplied the potential by 2, you will get an incorrect answer. And the point is for potentials, we don't multiply by this factor. To understand why, we can just look at how the standard free energy is defined. So, we know this relation between the standard Gibbs free energy and the standard potential. That is delta G0 is equal to minus NF E0, where F is the Faraday constant and N is this number of electrons. So, you can think of this product NF as the charge transferred. So, if you rearrange this, you get E0 to be minus delta G0 divided by NF. So, you can see here how when we calculate the potential, it is on a per charge transferred basis. So, the coefficient that we use to multiply is already accounted for in the denominator here. So, these values of E0 that are reported already include this factor, which is why we don't multiply this potential by 2 when we multiply this reaction by 2. So, now we have the cell reaction and the potential. So, let's see how we can construct a galvanic cell. Now, for that on the left side, we write down the oxidation reaction that is happening at the anode. Then we have the salt bridge, which is denoted by these two lines. On the right hand side, we write the reduction half reaction, which is happening at the cathode. So, we have this Fe3 plus aqueous going to Fe2 plus and since both of these are ions in solution, we have a solid platinum electrode to complete the circuit. And how I quickly check whether it is correct is at the right side, we write the reduction reaction and these three are written in alphabetical order. So, we have the anode, the bridge and the cathode. So, we know that we have written this down properly. And let's say if instead of these half reactions, we are given this galvanic cell and these reduction potentials, there is one more way to calculate the cell potential. So, we can write the cell potential as E0 cell, which is equal to the E0 of the cathode minus the E0 of the anode. So, in this case, it will be equal to 0.77, which is the cathode, minus the minus 0.76, which is at the anode. And notice that how when we use this method, the sign change that we saw here is taken care of in this formula itself. So, if we calculate this, we get 1.53 volts, which is the same that we got from this method. And so, we have the cell reaction, the cell potential and this is the galvanic cell.