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$$\ce{Cu^{2+} + Zn<=>Zn^{2+} + Cu}$$

$K_c=1.9*10^{37}$

$E_{cell}=1.1V$

I know that they both suggest that the forward reaction is going to completion but can someone please tell me a bit more about what both these values are suggesting and how these two values concord with each other.

Also, many books quote such reactions in only one direction and gives a value for $E_{cell}$. However, isn't $E_{cell}$ just another way of illustrating the condition of an equilibrium (i.e. in which direction the reaction would occur if left alone)... so how can a one way reation be shown

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  • $\begingroup$ See here for a derivation of the relationship between equilibrium constants and potentials. $\endgroup$ – Nicolau Saker Neto Jan 4 '14 at 14:55
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$E^\circ_{cell}$ and $K_c$ are related. Both indicate which direction of the reaction is spontaneous. Both tell you something about the position of equilibrium.

However, $E^\circ_{cell}$ is a directly measurable value. If you construct an electrochemical cell as using a zinc electrode immersed in a 1.0 M solution of say zinc chloride and a copper electrode immersed in a 1.0 M solution of copper (II) chloride with a salt bride of sodium chloride connecting the solutions, a voltmeter connected between the two electrodes will read the value of $E^\circ_{cell}$, at least initially. The voltage will change as the reaction progresses, but the initial voltageis $E^\circ_{cell}$ which is defined for standard conditions of 1.0 M concentration, 1 bar of pressure, and 25 $^\circ$C.

$$\ce{Zn|ZnCl2}(1.0 \text{ M})\parallel\ce{CuCl2}(1.0 \text{ M})|\ce{Cu} \ \ \ \ E^\circ_{cell}=1.1\text{ V}$$

The relationship between $E^\circ_{cell}$ and $K_c$ is the Nernst Equation. Note the use of both $E^\circ_{cell}$, the standard cell potential and $E_{cell}$, the observed cell potential under other conditions. The remainder of the alphabet soup is defined below. $$E_{cell}=E^\circ_{cell}-\frac{RT}{zF}\ln{Q}$$

  • $E^\circ_{cell}$ - Standard cell potential
  • $E_{cell}$ - nonstandard cell potential at other conditions
  • $R$ - the ideal gas constant (which appears here because it is equal to Avogadro's number times Boltzmann's constant
  • $T$ - the temperature in kelvins
  • $z$ - the number of moles of electrons being transferred in this redox process. If we wanted to use the number of individual electrons, then we would need to replace $R$ with $k_B$
  • $F$ - Faraday's constant
  • $Q$ - the reaction quotient

We need to use $Q$ in the Nernst equation, because this equation can relate the cell potential to any range of conditions, not just equilibrium.

At equilibrium $Q = K_c$ and $E_{cell} = 0 \text{ V}$ (there is no net change in the reaction composition, so there is no potential difference across the half-cells.

Thus, the Nernst Equation becomes:

$$0=E^\circ_{cell}-\frac{RT}{zF}\ln{K_c}$$ $$E^\circ_{cell} = \frac{RT}{zF}\ln{K_c}$$

To determine the values of $E^\circ_{cell}$ and $K_c$ for the reverse reaction...

For $E^\circ_{cell}$:

$$E^\circ_{cell} = -E^\circ_{rev}$$

For $K_c$

$$K_{rev} = \frac{1}{K_c}$$

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    $\begingroup$ Great post as always. Just a small thought regarding $z$. The quotient $\frac{RT}{zF}$ needs to have a dimension of volts, and by dimensional analysis, the $mol^{-1}$ from $R$ would cancel out with the $mol^{-1}$ from $F$, clearing the quotient of any dimension in quantity of matter. If $z$ had the dimensionality of moles, then it would reintroduce a unit that could not cancel out. I imagine that means the $z$ is only the adimensional stoichiometric coefficient for electrons in the reaction, and as such it can't be converted to number of electrons by substituting $R$ for $k_B$. $\endgroup$ – Nicolau Saker Neto Jan 5 '14 at 15:32
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    $\begingroup$ Rather, if $k_B$ is used, then I think that instead of $F$ it would be necessary to use $e$, the fundamental electric charge. $\endgroup$ – Nicolau Saker Neto Jan 5 '14 at 15:33

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