# Electrochemical Cells under Nonstandard Conditions

- Page ID
- 289

Electrochemical Cell Conventions cells are devices that use the transfer of energy, in the form of electrons, to measure the energy available after a given reaction. There are two forms of electrochemical cells: galvanic (voltaic) and electrolytic. Spontaneous reactions take place in galvanic cells and non-spontaneous reactions take place in electrolytic cells. Regardless of the resulting energy, each electrochemical cell consists of an anode, where oxidation takes place; and a cathode, where reduction takes place.

## Introduction

Anodes and cathodes are both called electrodes, and are two of the vital pieces in constructing an electrochemical cell. Electrochemical cells can take place under standard conditions or non-standard conditions (in both, electrons always flow from the anode to the cathode). Standard conditions are those that take place at 298.15 Kelvin (temperature), 1 atmosphere (pressure), and have a Molarity of 1.0 M for both the anode and cathode solutions. Non-Standard conditions occur when any of these three conditions is changed, but generally involve a change in concentration (check the Concentration Cell section for more details).

ANODE: oxidation --> always on the left

CATHODE: reduction --> always on the right

Recall:The Cell Potential is the potential (in volts) that results from a change in electron number. Cell potential if a cell at standard conditions can be obtained by the equation: E^{°}_{CELL} = E^{°}_{CAT} - E^{°}_{AN. }This can also be solved using the Standard Hydrogen Electrode.

Electrochemical reactions rarely occur under standard conditions. Even if we start at standard conditions, species involved in electrochemical reactions change in concentration throughout the reaction, removing them from standard conditions.

For electrochemical cells under non-standard conditions, we use the Nernst Equation:

E_{cell} = E^{°}_{cell} - [(RT)/(nF)]*ln Q

>E^{°}_{cell }= E^{°}_{cat - }E^{°}_{an }

>n = how many electrons were transferred between the cathode and the anode

>Q = activities (Q of homogenous or pure solids and liquids is 1; recall how to calculate this from concepts of equilibrium )

>R is the Ideal Gas Constant = 8.314 J/(mol K)

>F is Faraday's Constant = 96485 C/mol

As demonstrated by this equation, determining the electrochemical potential of electrochemical cells under non-standard conditions is almost identical to the process of finding the electrochemical cells under standard conditions. The difference, however, lies in the fact that another equation is used for reactions occurring under non-standard conditions because we take into account a change in concentration among the species.

Example 1 |
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Cu (s) l Cu 1. identify which species are reduce and which are oxidized. We know iron will be reduced (it's on the right of our cell diagram) and copper will be oxidized (it's on the left of our cell diagram) Cu → Cu Fe 2. write out the overall equation for the reaction (remember to multiply our equations with the appropriate numbers so the electrons cancel) 2Fe 3. find n (the number of electrons transferred) = 2 4. look at the reduction potential tables and solve E E° 5. Plug the standard electrode potential into the Nernst equation E E E E *note: since my E |

## Outside links

- www.compuchem.com/ecells.htm
- http://www.physchem.co.za/OB12-sys/batteries.htm
- http://www.chembio.uoguelph.ca/educm.../galvanic1.htm

## References

- Coleman, J.J. "Batteries, Electric." Wiley Interscience. 2000. 4 December 2008. <www.mrw.interscience.wiley.co...k/article.html>
- "Electrochemical reaction." Encyclopedia Britannica. 2008. Encyclopedia Britannic Online. 4 December 2008.< http://search.eb.com/eb/article-49338 >
- Petrucci, Ralph H.
__General Chemistry: Principles and Modern Applications 9th Ed.__New Jersey: Pearson Education Inc. 2007.

## Problem

Determine E_{cell}for the reaction in non-standard conditions:

Al (s) l Al^{3}^{+} (.36 M) ll Sn^{4+} (0.086 M), Sn^{2}^{+} (0.54 M) l Pt (s).

Also indicate which element is being oxidized and which element is being reduced as well as the anode and the cathode.

## Answer

Determine cell voltage:

Al(s)→ Al^{3}^{+}(aq) + 3e^{-} oxidation (anode) E°_{cell} = -1.676 V

E°_{cell }is the Standard REDUCTION potential for the equation written above. The voltage of the equation above is actually +1.676V since we would be looking at the standard OXIDATION potential (the equation above is an oxidation one).

Sn^{4}^{+} (aq) + 2e^{-} → Sn^{2}^{+} (aq) reduction (cathode) E°_{cell} = 0.154 V

The electrons need to be balanced: multiply the first reaction by 2 and the second reaction by 3, you should get the net equation to be:

2Al(s)+ 3Sn^{4}^{+} (aq) → 2Al^{3}^{+} (aq) + 3Sn^{2}^{+} (aq)

recall: E°_{cell }= E°_{cat }- E°_{an} (these are standard REDUCTION potentials), therefore E°_{cell }= 0.154 - (-1.676) = +1.830 V

Use the Nernst equation

E_{cell} = E^{°}_{cell} - [(RT)/(nF)] * ln Q

n = 6 (see oxidation-reduction equation, this is the number of electrons transferred)

E_{cell}= 1.830 - (8.314*298)/(6*96485) * ln ([Al^{3}^{+}]^{2}[Sn^{2}^{+}]^{3})/([Sn^{4}^{+}]^{3})

remember that solids are not included in Q

E_{cell}= 1.830 - (8.314*298)/(6*96485) log(.36M)^{2} (.54M)^{3} / (.086M)^{3 }= ^{ }**+1.851V (spontaneous because ****E _{cell} is **

**positive)**