Conc and Potential

When we calculate the standard potential we are getting a potential in volts for a cell made at standard conditions of concentration and pressure. This means that every aqueous species, both products and reactants, are at a concentration of 1 M. And, if there are gases, then those are all at 1 atm. Those are both pretty high concentrations and pressures in general terms. By changing to a new set of concentrations and pressures we can shift the standard potential voltage (\(E^\circ\)) to higher or lower values than the standard. This is completely handled via the Nernst equation.

\[E = E^\circ -{RT\over nF}\ln Q \]

Where the plain \(E\) is the non-standard potential. \(R\) is the universal gas constant, but this time we use the value of 8.314 J/mol K. \(F\) is the faraday constant (introduced in section 7.5) and is equal to 96485 C/mol e. And, a friendly reminder that \(Q\) is just the same equation as what the equilibrium constant (\(K\)) is - this was fully discussed at the end of section 6.4 here in chembook.

\[Q = {[\rm prod]^{coeff}\cdots\over [react]^{coeff}\cdots} \]

We can also just calculate what \(RT/F\) is (0.0257) and then do a math trick and switch to the more humam friendly log base10 to get

\[E = E^\circ -{0.05916\over n}\log Q \]

Best to Learn by Example!

Let's continue with our beloved Daniell cell of Zn | Zn2+ || Cu2+ | Cu which has a standard potential of +1.10 V and transfers 2e in the balanced equation. The overall reaction for this cell is

Zn(s) + Cu2+(aq) ⇌ Zn2+(aq) + Cu(s)

This means that \(Q\) will equal [Zn2+]/[Cu2+]. Remember we do not include pure solids and liquids in the expression. Now the Nernst equation for this cell reaction will be

\[E = +1.10 -{0.05916\over 2}\log{[{\rm Zn}^{2+}]\over [{\rm Cu}^{2+}]} \]

This allows us to get other potentials now by just putting in different concentrations of the two ions zinc(II) and copper(II). Let see what happens when the zinc(II) is lowered down to only 0.0010 M and the copper(II) pushed up to 2.0 M.

\[E = +1.10 -{0.05916\over 2}\log{0.0010\over 2.0} \]

\[E = +1.10 -(0.02958)\log(0.00050) \]

\[E = +1.10 -(-0.0976) = +1.10 + 0.0976\]

\[E = +1.1976 = +1.20 \;{\rm V}\]

Notice how lowering the concentration of the product and raising the concentration of the reactant increases the voltage. You can also do the exact opposite and the voltage will drop...

\[E = +1.10 -{0.08206\over 2}\log{2.0\over 0.0010} \]

\[E = +1.10 -(0.02958)\log(2000) \]

\[E = +1.10 -(+0.0976) = +1.10 - 0.0976\]

\[E = +1.0024 = +1.00 \;{\rm V}\]

Here we switched the two concentrations: raising the product and lowering the reactant - this results in the same abount of change (0.0976 V) but we now drop by that amount of voltage to 1.00 V. Notice this is very sensitive as well. We can measured voltage changes down to tenths of a millivolt (mV) very easily and cheap.

Using Voltage to Determine Unknown Concentration

We often purposefully design electrochemical cells such that the solution in one of the half-cells has an unknown concentration. We then measure the voltage and use it to calculate the concentration. This is how pH meters work.

Concentration Cells

A true concentration cell is driven purely by the differences in concentration of the same species but in two separate cells (anode / cathode). You can use any half reaction - you just need to use the SAME standard half reaction for both the oxidation and the reduction (reaction is just going 2 different directions). The easiest way to see this is to pick any half reaction with a metal and it's cation. Here is one with copper and copper(II) ion.

reductionCu2+(aq) + 2eCu(s) +0.34 V
oxidationCu(s)Cu2+(aq) + 2e –0.34 V
net rxnCu2+(aq) + Cu(s) Cu(s) + Cu2+(aq)   0.00 V

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