Write the shorthand notation for a galvanic cell that uses the following cell reaction. Include inert electrodes if necessary. 2 Fe1s2 + Cr2O72-1aq2 + 14 H+1aq2¡2 Fe3+1aq2 + 2 Cr3+1aq2 + 7 H2O1l2

Video transcript

Hello. Everyone in this video, we're considering this redox reaction right over here, we're trying to write a shorthand notation for this reaction and we can use an unearthly metal if necessary. So the standard sanitation is let's put this in black. So on the left side we're going to start off with the anodes than the A. Node ions than the cathode ions. And then of course our cathode. So this double line that put here, this denotes the phase boundary or the salt bridge. A standard sanitation is ours born from an 02 cathode as we did right over here. So to write the sanitation, we must first identify the cathode and the note. Now we can do that if we're able to determine the oxidation state of either nitrogen or manganese on both sides of the equation. So let's first try this out with our nitrogen. So taking the N. H four plus cannon that we have here know that hydrogen is always going to be plus one unless it's in a hydride compound with group one A or a to a medal. So then of course a nitrogen is unknown. But we do know for the higher regions and we have four items of this will provide this plus one charge. Of course this net charges just plus one. So simplifying this here, we get four equaling to just one. We can see that nitrogen. Then of course we're subtracting both sides by four and we get that N is equal to negative three. So the oxygen oxidation state of nitrogen is minus three on the reacting side And then on the other side here we see a N two value. So this is equal to zero because the oxidation state of an element in its natural state is going to be zero. So, we can see here that we're going from a negative three charge to zero. So we're increasing an oxidation number or a oxidation reaction. And that's when we have not just do this in red. We have two moles of RNH four plus to yield and to an Ask ASHA state. Since the economic an electrode from and to gas, we will have to use an inert electrode. We're gonna go ahead and use P. T. So our inert electrode, it's going to be platinum. Go ahead, scroll down here for a little more space. All right, So, the reduction reaction then is going to be our Mn of four minus To yield two moles of RMN 02 in its solid state. We cannot make an electrode with this over here since it's not a conductor. So, we can use an inert electrode here as well that we can use again, platinum. So are inert electrode will be platinum. So then our salutation then is our platinum Than our NH four plus. Then our end too. And we have this salt bridge here with our M N. 02 in a solid state than the M N 04 minus. And lastly the platinum in its solid state. So this right here is going to be my final answer for this problem.