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### Course: MCAT > Unit 9

Lesson 14: Electrochemistry- Electrochemistry questions
- Electrochemistry
- Redox reaction from dissolving zinc in copper sulfate
- Introduction to galvanic/voltaic cells
- Electrodes and voltage of Galvanic cell
- Shorthand notation for galvanic/voltaic cells
- Free energy and cell potential
- Standard reduction potentials
- Voltage as an intensive property
- Using reduction potentials
- Spontaneity and redox reactions
- Standard cell potential and the equilibrium constant
- Calculating the equilibrium constant from the standard cell potential edited
- Nernst equation
- Using the Nernst equation
- Concentration cell
- Introduction to electrolysis
- Quantitative electrolysis
- Electrolysis of molten sodium chloride edited
- Lead storage battery
- Nickel-cadmium battery

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# 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.

## Want to join the conversation?

- I thought to calculate Ecell you did Ered - Eox NOT Ered + Eox. I'm confused.(11 votes)
- 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.(6 votes)

- Since, we multiply the reduction reaction by two, won't the reduction potential also become twice?(9 votes)
- 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.(14 votes)

- If the Ecell doesn't increase with more moles , how do you get 12V?(3 votes)
- 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.(5 votes)

- I thought that E Cell was calculated like this: E Cell = (More Positive E Potential) - (Less Positive E potential)?(2 votes)
- 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.(2 votes)

- In the previous vedio [Standard Reduction Potential] at around0:49it 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(4 votes)

- 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.(2 votes)- Nope, both stay positive. If you had +1.30 for one and +1.20 for another element, then total overall voltage is the difference 0.10 and not 2.50. Btw, he does do this. It is the difference between 0.80 and -0.76, which is 1.56. In his example he calculates it as:

0.80 - (-0.76) = 0.80 + 0.76 = 1.56(1 vote)

- is Ecell=Ered+Eox the same as Ecell=Ecathode-Eanode?(2 votes)
- Whats the best way to pair the electrodes with electrolytes?(2 votes)
- why is Li strongest reducing agent and fluorine weakest?(2 votes)

## Video transcript

- [Voiceover] Let's say that
we're given this reaction and we're asked to find the
standard cell potential. We can do that by using our
standard reduction potentials. So if we look at this first half-reaction, we see that silver ion,
if you add an electron to a silver ion you get solid silver. So gain of electrons is reduction. So this is a reduction half-reaction. If we look at the reaction on the right, we can see that we are
reducing silver ions. So here's silver ion which is
being reduced to solid silver. So we need this half-reaction
as it's written. So this is going to be our
reduction half-reaction. So we have Ag plus, so we have Ag plus plus an electron giving us a solid silver here. The standard reduction
potential for this half-reaction is positive .8 volts. So this is positive .8 volts. Next, let's look at
what else is happening. We're turning solid zinc
into zinc two plus ions. How do you turn solid
zinc into zinc two plus? Solid zinc would need to lose two electrons to turn into zinc two plus. Loss of electrons is oxidation. So if we go over here
to this half-reaction, we can see that this
half-reaction is written as a reduction half-reaction. Zinc two plus is being reduced,
is gaining two electrons, to turn into solid zinc. We need to write this as
an oxidation half-reaction. So it's given to us as
a reduction, we need to write it as an oxidation half-reaction, we simply need to reverse it. So we need to start with solid zinc. So we start with solid zinc here. And for solid zinc to turn
into zinc two plus ions, we need to lose two electrons. So loss of electrons is oxidation. So we need to find the standard
oxidation potential next. And we can do that by looking
at our table once again. The standard reduction potential for this half-reaction as it's written
is negative .76 volts. Since we reversed this half-reaction, we reversed this half-reaction, we need to change the sign
from negative to positive. So for the reduction
potential it's negative .76, for the oxidation potential,
it's positive .76. So the standard oxidation
potential for this half-reaction is positive .76 volts. Next, we need to look at moles. So if we look at our reaction, we have two moles of silver ions being reduced to two
moles of solid silver. Down here we have one mole of silver ions being reduced to one mole of solid silver. So we need to multiply
everything by two here, so we get two moles of silver ions, which require two moles of electrons, to turn into two moles of solid silver. Now it might be really tempting to say, oh, well then don't we just
multiply our reduction potential by two as well? And we don't, for the
reason that we talked about in the previous video. Voltage is an intensive property. So it doesn't matter if you're forming one mole of silver or two moles of silver, the voltage is the same. So we're gonna leave
the voltage at .8 volts. Alright, next we have the
number of electrons equal, so we can add our two
half-reactions together to get our overall reaction. So if you add the reduction half-reaction and the oxidation half-reaction, together we get our overall reaction. So the electrons lost by zinc, these two electrons here, are the same electrons that are gained by the silver ions. So those cancel out. And on the left side we
would have two Ag plus plus solid zinc. So let's write that in here. Two Ag plus, plus solid zinc, and for our products, so over here on the right, we have two Ag plus zinc two plus. So we have two Ag plus zinc two plus. Notice this is our reaction. So if I box this here, this overall reaction
that we've just found, is the same as what we
were given in our problem. So we found our overall reaction. But remember, they wanted us to find the standard cell potential. So how do we find the
standard cell potential? We talked about how to do
this in an earlier video. To find the standard cell potential, you can think about how we
found the overall reaction. We added the reduction half-reaction and the oxidation half-reaction together to get the overall reaction. So to find the standard cell potential we need to add the standard
reduction potential and the standard oxidation
potential together. If we add the standard reduction potential and the standard oxidation
potential together we should get the standard
potential for the cell. So let's go ahead and do that. The standard cell potential is equal to, this would be positive .8 volts. So positive .8 volts plus positive .76 volts. Plus positive .76 volts. Which is equal to positive 1.56 volts. So the standard cell potential
is positive 1.56 volts. Remember, a positive
value for your potential means a spontaneous reaction. So this is how to use
your reduction potentials to figure out a standard cell potential.