How to use a table of standard reduction potentials to calculate standard cell potential. Identifying trends in oxidizing and reducing agent strength. Created by Jay.
Want to join the conversation?
- Reactivity is often described by electronegativity or by electrodpotential. What is the difference? Eg., halogenes have the most positive standard potential therefore are the most reactive elements. Also, the EN values are high for halogenes, so they bond with other elements easiliy. Do EN and elektrodepotential donote the same thing basically, describing to what extent an atom can attract electrons? Or should they be used in different contexts?(24 votes)
- This is my opinion but not sure though. Electronegativity is how strongly the element hogs the election ONCE the covalent bond is made. Electropotential is the tendancy of the element to lose/gain electrons (so I see it close to ionization energy/electron affinity definitions). Nevertheless, they are related.(2 votes)
- lithium despite having highest ionisation potential how it behaves as a strong reducing agent?(4 votes)
- Before answering to your question we need to know that :
1) all the reducing agents undergo oxidation themselves
2) and we also know that oxidation is the loss of electrons and that reduction is gain of electrons
3) ionistasion potential determines the ease with with an element can lose an electron
So coming to your question ..
as we know lithium has high ionisation potential ,. i.e it can lose electrons with much ease than any other metal and we also know that reducing agents lose electrons( the easier they lose electrosn the stronger reducing agents they are)and that lithium can lose electrons with greater ease ..hence it is strong reducing agent.
Hope it helps..if any question feel free to comment .. :)(15 votes)
- How is it that 2H+ has a higher reduction potential than Zn++ but a lower reduction potential than Cu++? I see that H is more electronegative than Zn and Cu, and H has a higher ionization energy than Zn and Cu (both suggesting H should have a higher reduction potential than Zn and Cu). I'd think there should be a direct inverse correlation between the reduction potential (the tendency of the cation to gain an election) and ionization energy (the energy needed for the neutral atom to release an electron). Does the answer lie in the fact that we're comparing TWO H+'s with a single Cu++ and and single ZN++ ?(4 votes)
- Hi Vance, I had exactly the same question. From some other comments (below) I noted that it is not only the electronegativity that counts but also what happens with the metal ions in water. A possible answer could thus be that there is a difference in the hydration energy between Zn++ and Cu++ so that Zn++ actually acts as a reducing agent against hydrogen while Cu++ acts as an oxidation agent. Why such a difference occur, I do not know as the size of the two ions should be fairly similar so I am not sure that this is a correct explanation.(2 votes)
- Is it possible for zinc and lithium to be in a redox reaction? if so would the zinc ion reduce and the lithium oxidise as the reduction potential of -0.76V (zinc) is more positive than -3.05V(lithium)?(4 votes)
- Okay!But from what i have studied the formula for E(cell)=E(cathode) - E(anode) How would you define that? and beside the topic is of Standard Reduction Potential so why are we talking about oxidation?? my guess is that the formula above i gave is more suitable?thoughts??(1 vote)
- Usama, your confusion is perfectly understandable! Let me try to help...
The formula you have proposed is perfectly valid. Recall from the video that all the potentials in the chart were referring to reduction potential. The higher and more positive the reduction potential the greater likelihood that species would be reduced. Also, remember that the cathode is the site of reduction so its reduction potential will be positive (likely to spontaneously reduce), and that the anode is the site of oxidation so its REDUCTION potential will be negative (not likely to be reduced). In the formula E(cell)= E(cathode) - E(anode) we have a positive value for the E (cathode) and a negative value for E(anode) since their REDUCTION potentials are positive and negative, respectively. Plugging in a positive value at E(cathode) and a negative value for E(anode) leaves us with an overall positive E(cell), which correlates to a spontaneous reaction [delta G= -nFE(cell)].
This is good news! In order to have a spontaneous reaction, as is the case in voltaic cells, we want a species that is easily reduced (positive reduction potential) to be able to gain its electrons from a species who is easily oxidized (negative reduction potential). This is exactly what the equation in your question is stating. If you have a highly positive reduction potential at the cathode and a highly negative reduction potential at the anode then plugging the terms in to the equation leads to a very spontaneous reaction. Voila!
You may confused as to why in the video the speaker says you can just add the E(reduction) to E(oxidation) and get E(cell). Well, don't be! E(reduction) is the same as E(cathode) and E(oxidation) is the same as -E(anode)....Remember that E(anode) is the REDUCTION potential at the anode, thus if we switch it's sign by multiplying it by a negative number we automatically get it's OXIDATION potential or E(oxidation). Taking your equation E(cell)= E(cathode)-E(anode) and replacing -E(anode) with E(oxidation you are left with E(cell)= E(cathode)+ E(oxidation).(7 votes)
- Is it theoretically possible to have two electrodes made of the same material (i.e. Zinc)?
Each would be an anode and a cathode depending on the electrolyte solution used. Keeping the example of Zinc, one could use maybe a solution that contains sulphate ions for the case of the anode, and diluted Zn2+ ions for the cathode.
If this worked, would the voltage of the cell be equal to 1.52 (0.76 V + 0.76 V)?(3 votes)
- How would you compare Zn 2+ and Cu? which is the better oxidizing/reducing agent(2 votes)
- How is the standard reduction (half-reactions) calculated? in my chemistry class, they are simply given to us in a chart, but I'm curious the specific calculation chemists go through to calculate each one of the reduction half-reaction potentials.(2 votes)
- Scientists create a galvanic cell in which one half-cell contains the half-reaction to be determined.
The other half-cell is a standard hydrogen electrode (SHE) in which the half-reaction is:
2H⁺ + 2e⁻ = H₂
E° for the SHE is defined as 0.00 V.
Then E° for the half-reaction is the same as the measured value of E° for the cell.
The values in your chart were measured against a SHE.(2 votes)
- How come Li has more reduction potential than Na or K? Isn't the valence electron in Li more strongly attracted to the nucleus (due to the shorter distance) than the valence electrons in Na and K? K is after all more reactive than Li, doesn't that imply a greater reducing potential?(0 votes)
- Reduction = gain in electrons. K would find it harder to gain electrons than Li because there are more electron shells around K. This means that atomic radius has increased, so the e- are held with less attraction to the nucleus. Increased sheilding also contributes. This is why K is so reactive - it's electrons are held so losely they easily break away. In reduction you want to opposite to happen, a gain in electrons, so a strong force of attraction on your valency electrons is needed. Li has the smallest radius (and all of group one has the same nuclear charge) so Li has the strongest force of attraction and thus most easily atracts more electrons.(3 votes)
- what is the difference between EMF and potential difference?(1 vote)
- EMF is a SOURCE of potential difference.
There's a potential difference across a current-carrying resistor, but you can't use it as an energy source, right?(2 votes)
- [Voiceover] Here we have a table of Standard Reduction Potentials, and this is a shortened version, but you can see on the left side, we have different half-reactions. All of these half-reactions are written as reduction half-reactions. Remember gain of electrons is reduction. If we look at our first half-reaction, we have silver ion gaining an electron to form a solid silver. That's a reduction half-reaction. The standard reduction potential turns out to be +.80 volts. That's compared to this half-reaction down here, which corresponds to the standard hydrogen electrode, which is the reference value. This has a potential of zero volts. All of our other half-reactions are compared to this one. The more positive the value is for the standard reduction potential, the more likely the substance is to be reduced. Let's compare the reduction of copper 2+ ions to the reduction of zinc 2+ ions. Let's compare these two half-reactions. If we are reducing copper 2+ to solid copper, the standard reduction potential is +.34 volts. If we are reducing zinc 2+ to solid zinc, the standard reduction potential turns out to be -.76 volts. The more positive value, the more likely the substance is to be reduced, so obviously +.34 is more positive than -.76. We know that this must be the reduction half-reaction. If we're talking about a redox reaction involving copper and zinc, this must be our reduction half-reaction. Let's go ahead and write that. This is our reduction half-reaction where we have copper 2+ ions gaining two electrons to turn into solid copper. The standard reduction potential is +.34 volts. This is equal to +.34 volts. We know in a redox reaction something is reduced and something is oxidized, and since we already have our reduction half-reaction, we need an oxidation half-reaction. That must mean that this should be our oxidation half-reaction, but here we have it written as a reduction half-reaction, so we need to reverse this reaction to show it as an oxidation. We need to start with solid zinc on the left side. We start with solid zinc on the left side, and zinc is oxidized into zinc 2+ ions, and we're losing two electrons. Remember loss of electrons is oxidation. Now this is an oxidation half-reaction. We need to find the standard oxidation potential for this half-reaction. We can do that by looking at our table here. So -.76 is the standard reduction potential. Since we reversed our half-reaction, we just need to change the sign. The oxidation potential must be +.76. All we need to do is reverse the sign to get our standard oxidation potential, so we get +.76. To find our overall redox reaction, we just need to add together our two half-reactions. To find our overall redox reaction here, we add the reduction half-reaction and the oxidation half-reaction. These electrons would cancel out, and on the left sides we would get copper 2+ ions. This would be copper 2+ ions and solid zinc. On the right side for our products, we would get solid copper and zinc 2+ ions in solution. We get solid copper and zinc 2+ ions. This overall redox reaction should look very familiar to you because this is the spontaneous redox reaction that we've talked about in the last several videos as our example of a voltaic cell, all right? We know this is a spontaneous reaction. How do we find the potential for the cell, all right? How do we find the standard cell potential? How do we find the potential for the entire cell? To find the overall reaction, we add together our reduction half-reaction and our oxidation half-reaction. That gave us our overall reaction. To find our standard cell potential, we just need to add together our reduction potential for the half-reaction and the oxidation potential for the oxidation half-reaction. To find the potential for the cell, we add the reduction potential and the oxidation potential. We get when we do that, we're gonna get +.34 volts is the potential for the reduction half-reaction, and +.76 volts is the potential for the oxidation half-reaction. That gives us our standard cell potential. For our cell the potential is equal to +1.10 volts, which we already know this from previous videos, right? I talked about the fact that you can use a voltmeter to measure the potential difference, to measure the voltage of a voltaic cell. You're gonna get a +1.10 volts under standard conditions. That's one of the nice things about the standard reduction potential table. We can calculate the voltage of our voltaic cells this way. Let's look in more detail at our half-reactions. Let's start with the oxidation half-reaction. We know that zinc is being oxidized, right? Zinc is losing 2 electrons, and those two electrons that zinc loses are the same two electrons that caused the reduction of copper. Zinc is the agent for the reduction of copper. We say that zinc is the reducing agent. Sometimes students find this confusing because zinc is being oxidized, so why is it the reducing agent? Zinc is the agent for the reduction of something else, in this case, copper 2+ ions. So zinc is the reducing agent. Copper 2+ is gaining those two electrons, so copper 2+ is being reduced, but because copper 2+ is gaining those two electrons, it allows zinc to be oxidized. Copper 2+ is the agent for the oxidation of zinc. Copper 2+ is our oxidizing agent for our redox reaction. Let's look at our standard reduction potential table and let's see if that can help us understand oxidizing agents and reducing agents. We've been comparing these two half-reactions, right? These two half-reactions right here. Let's compare copper 2+ ions to zinc 2+ ions, right? Copper 2+ we know is more easily reduced, right? It has the higher, has the more positive value, I should say, for the standard reduction potential. Copper 2+ is more easily reduced, and therefore, copper 2+ is a stronger oxidizing agent than zinc 2+. As you go up on your standard reduction potential, you're increasing in the tendency for something to be reduced, and therefore, you're increasing the strength as an oxidizing agent. As you move up on your standard reduction potential, increased strength as an oxidizing agent. Copper 2+ is a stronger oxidizing agent than zinc 2+. All right, let's think about the opposite. As you move down on your reduction potentials, as you move down here, let's compare solid copper and solid zinc, right? We know that solid zinc was our reducing agent in our reaction, and that's because the reduction potential was the more negative one. That means this is more likely to be the oxidation half-reaction. As you move down on your reduction potential, you have an increasing tendency to be oxidized. Therefore, you have an increasing strength as a reducing agent. Zinc, right, is a stronger reducing agent than copper because, again, looking at the reduction potentials, you know that it's more likely to be oxidized. Going down on your reduction potentials, increased tendency to be oxidized, therefore, increased strength as a reducing agent. If you look at lithium, right? Lithium here is even more negative for the reduction potential. Therefore, this is even more likely to be an oxidation half-reaction. Lithium is a stronger reducing agent than zinc.