Cell Potential: The Nernst Equation - Video Tutorials & Practice Problems
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concept
The Reaction Quotient
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Now before we can talk about sub potential and its connection to the Nernst equation, it's important to revisit the idea of our reaction quotient. Now recall that the reaction quotient which uses the variable q is a ratio of product to reacting concentrations at a particular time, and we're gonna say it can be calculated by setting up an expression and ignoring solids and liquids. Remember, like your equilibrium constant k, we only care about gaseous and aqueous species. Now how does this relate to electro count? Well, we're going to say here that for electrochemical cells, it helps to find the max potential at the exact moment the cell circuit is connected. Because as we know, over time our voltage is going to decrease as our battery, as our electrochemical cell overall degrades. Right? So this is where the reaction quotient comes into play. It helps us to find that exact moment in which our voltage is going to be at its highest.
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example
The Reaction Quotient Example
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Here it says, what is the reaction quotient for the following redox reaction with the given concentrations? Alright, so here we have lead 2 ion reacting with 2 moles of potassium solid to produce 1 mole of lead solid plus 2 moles of potassium ion. Here we're going to say that the concentration of let 2 ion and potassium ion are given as these molarities. Now remember that your reaction quotientqcanbe solvedbysettingupitsexpression, andq equals products over reactants. Now remember with this expression we ignore solids and liquids, so the solids we're going to ignore. So it basically becomes the concentration of k positive and remember because there's a 2 here that becomes our exponent, so k positive squared divided by pv2plus. Here we plug in the values for each one. So for potassium ion, it's 0.0015, which will be squared divided by 0.0880. When we punch this into our calculators, we get 2.6 times 10 to the negative 5. So this represents the value for our reaction quotient q.
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concept
The Nernst Equation
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Now recall that our standard cell potential is calculated when ions in half cells have values of 1 molar, 1 atmosphere, and a pH equal to 7. Now the Nernst equation is used to find the cell potential when ion concentration or concentrations do not equal 1 molar. So that's when we're going to use our Nernst equation, because it's going to help us to calculate nonstandard cell potentials. Now here, the Nernst equation formula is, we're gonna say e cell, e sub cell here represents our nonstandard cell potential, and it equals our standard cell potential, which is e naught subcell, minus 0.05916 over n times log of q. N equals the moles of electrons that are transferred between our species within our redox reaction, and q is just our reaction quotient. So here this represents our Nernst equation formula. Again, it helps us to determine our non standard cell potentials when ion concentrations are not equal to 1 molar.
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example
The Nernst Equation Example
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Here it says to calculate the cell potential for reaction at 25 degrees Celsius, we're given the following ionic concentrations and standard reduction potentials. Here, we're given the overall redox reaction as this, and we're given the concentrations of our ions of cobalt 3 ion and magnesium ions as 1 molar and 0.0033 molar respectively. In addition to this, we're given the standard reduction potentials in the form of these half reactions. Now notice they ask us to determine cell potential, not standard cell potential, there is a difference. So to find our non standard cell potential, we use the Nernst equation. So that's going to be cell potential, which is E sub cell, equals standard cell potential, which is E naught sub cell minus 0.05916 over n times log of q. Here we're going to find out what q is first. Remember q equals products over reactants, It ignores solids and liquids. So here it would equal mg 2+ cubed because of the 3, divided by c03+quared because of the 2. Here we plug in the values, so this is 0.0033 cubed divided by 1 molar squared. Here that gives me a q of 3.5937 times 10 to the negative 8. Here I'm not rounding until I get my final answer at the end. So we find out what q is. Now, what do we have to find next? Let's try to find out what n is. N is the number of multiple electrons transferred. Remember, the multiple electrons have to match in both half reactions. This one here has 3 and this one here has 2. Their lowest common multiple is 6. Okay? So it's 6 electrons that have been transferred. This would also explain our coefficients here. Those coefficients came into play because we had to multiply this by 2 to give us 6 electrons, and we had to multiply this by 3 to give us 6 electrons. Here this will be log of 3.59 37 times 10 to the negative 8. Now remember multiplying your half reactions does nothing to our standard reduction potentials, They stay those numbers because those values are based on the identity of the elements being reduced, not on the amounts of them. K? So I can multiply these by a million. These reduction potential would stay the same. So we found out almost everything, the only piece that's missing is our standard reduct is our standard cell potential. Remember, standard cell potential which is e's e naught subcell equals cathode minus anode. Cathode, remember, that's a site of reduction. Anode is a site of oxidation. If we look at the overall redox reaction, that'll help us determine what's been oxidized and what's been reduced. Cobalt 3 has an oxidation number of plus 3 because its charge is plus 3, and it goes to 0. Its oxidation number has been reduced, Therefore, it is the cathode. Magnesium goes from 0 to plus 2, its oxidation number increased, so it's been oxidized. So now we can do cathode minus anode. So that'd be cathode is 1.82 volts minus a minus 2.37 volts. Now remember a minus of a minus really means that we're adding them together. So that comes out to 1.82 volts plus 2.37 volts. So that comes out to 4.19 volts. That would be our standard cell potential. So now we do 4.19 volts minus 0.05916 divided by 6 times log of our q value. When you punch this into your calculator, you will you'll get back 4.26 volts. So this would represent our regular non standard cell potential, so that would be our final answer.
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Problem
Problem
If [Br–] = 0.010 M and [Al3+] = 0.022 M, predict whether the following reaction would proceed spontaneously as written at 25ºC:
Al (s) + Br2 (l) ⇌ Al3+ (aq) + Br– (aq)
Standard Reduction Potentials
Al3+ (aq) + 3 e– → Al (s) E°red = –1.66 V
Br2 (l) + 2 e– → Br– (aq) E°red = +1.09 V
A
2.90 V
B
2.60 V
C
3.03 V
D
2.75 V
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Problem
Problem
Determine [Fe2+] for the following galvanic cell at 25ºC if given [Sn2+] = 0.072 M, [Fe3+] = 0.0219 M, and [Sn4+] = 0.00345 M.