9
$\begingroup$

A $\ce{Cr(s) | Cr^3+ || Cd^2+ | Cd(s)}$ galvanic cell has redox half reactions:

$$ \begin{align} \ce{Cd^2+ + 2e- &-> Cd(s)} &\quad E^\circ &= \pu{-0.4 V}\tag{R1}\\ \ce{Cr^3+ + 3e- &-> Cr(s)} &\quad E^\circ &= \pu{-0.7 V}\tag{R2} \end{align} $$

Balanced reaction equation then is

$$\ce{3 Cd^2+ + 2 Cr(s) -> 2 Cr^3+ + 3 Cd(s)}\tag{R3}$$

$$E_\mathrm{cell} = E_\mathrm{red} + E_\mathrm{ox} = \pu{-0.4 V} + \pu{0.7 V} = \pu{0.3 V}$$

Why is $E_\mathrm{cell}^\circ$ independent of molar coefficients $(3,2)?$

$\endgroup$
4
$\begingroup$

You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions. Standard thermodynamic conditions are usually

  • A temperature of 298 K
  • A pressure of gaseous components of 1 atm (or 1 bar)
  • A concentration of 1 M

The reference reduction potentials you quote are all standard potentials. These are the contributions to the cell potential when the half-cells meet the criteria above.

$E$ is the cell potential at all other conditions. The correction of $E^\circ$ to $E$ involves temperature, pressure, concentration, and stoichiometry. Behold the Nernst equation:

$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$

$\endgroup$
3
  • $\begingroup$ Who is the god particle? $\endgroup$ – Buraian Jun 7 at 11:29
  • $\begingroup$ What Q should we take? We can scale up the reaction by any factor and Q would be different $\endgroup$ – Buraian Jun 7 at 11:39
  • $\begingroup$ @Buraian - You take whatever Q reflects your current situation. The Nernst equation allows you to calculate the cell potential given any starting concentrations in the half cells. $\endgroup$ – Ben Norris Jun 7 at 12:36
2
$\begingroup$

The standard electrode potentials are measured in their standard states when the concentration of the electrolyte solutions are fixed as $1M$ and temperature is $298K$. However, in actual practice electrochemical cells don't have always fixed concentrations of the electrolyte solutions. The electrode potentials depend on the concentration of the electrolyte solutions. Nernst gave a relationship between electrode potentials and the concentration of the electrolyte solutions known as Nernst equation. Even the electrode potential which you have posted (assuming they are not standard potentials) is obtained by Nernst equation, which shows the dependence of electrode potential on concentration (in other words molar coefficients).

$\endgroup$
2
$\begingroup$

Why is E$^\circ_\mathrm{cell}$ independent of molar coefficients (3,2)?

The cell potential is the potential an electron experiences; the coefficients of the equation have no bearing on this. When you use the cell potential to calculate the electrical work a cell does, however, you would multiply by the charge that is flowing through the wire, which would involve $z$, the number of electrons transferred.

So if we take the chemical equation as written by the OP,

$$\ce{3 Cd^2+ + 2 Cr(s) -> 2 Cr^3+ + 3 Cd(s)}\tag{one batch}$$

and multiply it by two, we would have

$$\ce{6 Cd^2+ + 4 Cr(s) -> 4 Cr^3+ + 6 Cd(s)}\tag{two batches}$$

This would not change the voltage observed between the electrodes. However, the molar Gibbs energy for the reaction would be different by a factor of two because the relationship between cell potential and Gibbs energy includes $z$, the number of transferred electrons per reaction:

$$\Delta_r G = -z F E_\mathrm{cell} $$

This is because the molar Gibbs energy of reaction is based on the extent of reaction (which itself is dependent on which coefficients you choose for the reaction, i.e. one batch or two batches in the example).

On the other hand, if we calculate the work available from oxidizing a given mass of elemental chromium in this reaction, we get the same answer no matter which equation we base it on (as it should be). Finally, having more chromium react does not change the voltage of the cell (as long as the solutes are maintained at constant concentration).

$\endgroup$
0
$\begingroup$

When we scale a half reaction in order to balance the number of electrons with the other half reaction, we do not change the intrinsic joules per coulomb (Volts) of the electrons.

So Fe2+ +2e- -------> Fe (s) E = -0.44V

means that the 2 electrons carry -0.44 Joules per coulomb of energy

If we double the reaction, the reaction energy is doubled, but so is the number of electrons. So the J/C stays the same. That is the standard reduction potential remains -0.44V

So 2Fe2+ +4e- -------> Fe (s) E = -0.44V

But when we are trying to combine the 2 reduction reactions as in the case of:

Fe2+ +2e- -------> Fe (s) E = -0.44V

Fe3+ +e- -------> Fe2+ (s) E = +0.77V

to get the standard reduction potential for :

Fe3+ +3e- -------> Fe (s) E = ??

this is a different concept. Here we are not scaling half reactions in order to combine them into a single full reaction. We are trying to combine 2 half reactions to make a new half reaction (with electrons still present in the equation)

At first glance it's tempting to just add the potentials, since adding the half reactions does give the desired half reaction :

  1. Fe2+ +2e- -------> Fe (s) E = -0.44V
  2. Fe3+ +e- -------> Fe2+ (s) E = +0.77V

Fe2+ +2e- Fe3+ +e- -------> Fe (s) Fe2+ (s) E = ?

(now cancel the Fe2+ on each side)

  1. Fe3+ +3e- -------> Fe (s) E = ?

But we cant just add the potentials. This is still a half reaction, so the potential is all about the energy per electron in Joules /Coulomb (V)

In reaction 1) the 2 electron carry -0.44 Joules per coulomb, and in reaction 2) the 1 electron carries 0.77 Joules per coulomb

So to get the J/C for reaction 3) we need the weighted average J/C of the 3 electrons

So that 2/3 electrons at -0.44 plus 1/3 electrons at +0.77

2/3 * (-0.44) + 1/3 * (0.77) = =0.037 Joules per coulomb (volts)

Hope that helps.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.