Effect of Concentration Changes on Cell Potential

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I The Nernst Equation Reported cell potentials are typically measured under standard conditions:

Solutes in aqueous solutions have a concentration of 1.0 mol dm–3 Gaseous reactants or products have a pressure of 1 atm Measurements are taken at 298 K

Galvanic cells seldom operate under standard conditions. Ecell measured under non-

standard conditions is not equal to (E o— )cell, the cell potential measured under standard conditions.


It has been determined that cell potentials are related to concentrations of reactants and products, and to temperature via the Nernst equation, as follows:

Ecell = (E o— )cell –

€

RTnF

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ ln Q

where R is the gas constant (8.31 J K–1 mol–1) T is the temperature (K) n is the number of moles of electrons

transferred between oxidising and reducing agents

F is the Faraday’s constant (96500 C mol–1) Q is the reaction quotient, which is based on

the cell reaction

This equation gives the Ecell measured under non-standard conditions. The Nernst equation is sometimes expressed in terms of base 10 logarithm. For a

cell at 298 K, the above equation becomes:


€

0.0592n

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10 Q


II Using the Nernst Equation to Illustrate the Effect of Dilution on (E o— )cell Consider the following galvanic cell set up at 298 K:

Al(s) | Al3+(aq) || Ni2+(aq) | Ni(s)

Anode : Al(s) → Al3+(aq) + 3e– Cathode : Ni2+(aq) + 2e– → Ni(s) Overall cell reaction : 2Al(s) + 3Ni2+(aq) → 2Al3+(aq) + 3Ni(s)

Under standard conditions, [Al3+] and [Ni2+] are both 1.00 mol dm–3,

(E o— )cell = (E o— )reduction – (E o— )oxidation = –0.25 – (–1.66) = +1.41 V


Scenario 1: How does (E o— )cell change when water is added to the Ni2+/Ni half cell such that [Ni2+]

is decreased ten times? (All measurements are taken at 298 K.)


€

0.0592n

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10 Q

= +1.41 –

€

0.05926

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10

€

[Al3+]2[Ni2+]3

= +1.41 –

€

0.05926

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10

€

11

10⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

3

= +1.41 – 0.0296 = +1.38 V

Note: Six electrons are tranferred during the oxidation of Al and the reduction Ni2+. Thus, n = 6.

(Q > 1) Ecell decreases.


Scenario 2: How does (E o— )cell change when water is added to the Al3+/Al half cell such that [Al3+]

is decreased ten times? (All measurements are taken at 298 K.)


€

0.0592n

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10 Q

= +1.41 –

€

0.05926

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10

€

[Al3+]2[Ni2+]3

= +1.41 –

€

0.05926

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10

€

110⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

2

1

= +1.41 – (–0.0197) = +1.43 V

(Q < 1) Ecell increases.


Scenario 3: How does (E o— )cell change when water is added to both half cells such that [Al3+] and

[Ni2+] are each decreased by ten times? (All measurements are taken at 298 K.)


€

0.0592n

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10 Q

= +1.41 –

€

0.05926

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10

€

[Al3+]2[Ni2+]3

= +1.41 –

€

0.05926

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟ log10

€

110⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

2

110⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

3

= +1.41 – 0.00987 = +1.40 V

(Resulting Q > 1) Ecell decreases, albeit slightly.


Conclusion In general, any change to the cell that increases Q decreases Ecell, while any change that decreases Q will increase Ecell. Thus, adding reactant or removing product increases Ecell. while removing reactant or adding product decreases Ecell. In the above example, when reactant concentration ([Ni2+]) is smaller than 1 mol dm–3, Ecell is less than

(E o— )cell (Scenario 1). when product concentration ([Al3+]) is smaller than 1 mol dm–3, Ecell is more than

(E o— )cell (Scenario 2). If concentrations of both products and reactants are decreased in the cell, whether

Ecell is more or less than (E o— )cell depends on the resulting Q: Q > 1, Ecell is less than (E o— )cell (Scenario 3) Q = 1, there is no change to (E o— )cell, i.e. Ecell = (E o— )cell Q < 1, Ecell is more than (E o— )cell

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Effect of Concentration Changes on Cell Potential