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### Course: Chemistry library>Unit 14

Lesson 3: Standard cell potentials

# Using reduction potentials

Learn how to calculate the standard cell potential using standard reduction potentials. This video explains the process of oxidation and reduction, how to identify half-reactions, and the role of moles in these reactions. Created by Jay.

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• I thought to calculate Ecell you did Ered - Eox NOT Ered + Eox. I'm confused.
• Ecell=Ered+Eox is correct. But you can also calculate it like this:
Ecell=Ered(Cu+2/Cu)-Ered(Zn+2/Zn) an example using Daniell cell, because in this new ecuation you are acknowledging that the Zn is being oxidized.
• Since, we multiply the reduction reaction by two, won't the reduction potential also become twice?
• Nope! That's a great question, since this is a common point of confusion when learning about reduction potential. The reduction potentials have units of volts, and are defined per electron for that half reaction. Even if you multiple the reduction reaction by two, you haven't changed the underlying half reaction, and the energy per electron (and thus the reduction potential) doesn't change.
• If the Ecell doesn't increase with more moles , how do you get 12V?
• Ion concentrations in the solution have a strong effect on the voltages across the metal/solution boundary. Alternatively, since you don't want moles, have a look at the Nernst equation which will show two other variables (others are constant):
z the electrons transferred per mole
T the absolute temperature
(1 vote)
• hi
I really want to understand what the purpose of finding the E-cell is please can you help me?
(1 vote)
• If you think about the practical application of this sort of thing, we are interested in the Voltage difference across cathode and anode, because we can use this potential difference (deltaV) to power things. This is the idea behind a battery. If you create a 12 V battery (or a cell with and E-cell of 12 V) then you can use that potential difference to power devices. A combination of any two oxidizing and reducing agents gives a certain E-cell as calculated in the video. This tells you what the Voltage difference will be between the cathode and anode. Thus, you get information about the ability of the battery.
• I thought that E Cell was calculated like this: E Cell = (More Positive E Potential) - (Less Positive E potential)?
• The equation that you give 'E Cell = (More Positive E Potential) - (Less Positive E potential)' only works if the less positive E potential is the reducing agent and the more positive E potential is the oxidizing agent, when the reducing agent is negative, and you haven't switched the sign. this 'E Cell = (More Positive E Potential) - (Less Positive E potential)' works only in these select cases.
• In the previous vedio [Standard Reduction Potential] at around it is said the more positive the value is for the standard reduction potential, the more likely the substance is to be reduced. As reduction occurs in the cathode the electrode is cathode according to the last sentence. Now, here if we do it, the oxidation reaction is Zn to Zn2+ where Eox=+0.76V [since we are reversing the reaction given in the table as it is oxidation] and the reduction reaction is 2Ag+ to 2Ag where Ered=+0.80V. Now if we calculate Ecell=Ecathode[i.e. Ered]-Eanode[i.e. Eox] it's comming to +0.04V. Now maybe I am doing a mistake but technically I am confused. Why aren't we taking Eox as +0.76V i.e. why are we taking the negative value?
(1 vote)
• Ecell = Ecath - Eanode when both of these are 'reduction' potentials, and Ecell = Eredn + Eoxdn when you've multiplied E(redn)anode with -1.

So you either use the first convention, or the second. You don't mix the two.

Ecell = E(redn)cathode - E(redn)anode = 0.80 - (-0.76) = 1.56

OR

Ecel = E(redn) + E(oxdn) = 0.80 + 0.76 = 1.56
• Supposing I had two half reactions that both had +ve reduction potentials. If one of them had to be converted into a reduction half reaction, would I have to make the value negative?
Although I notice he doesn't do it for Ag.