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I    The Nernst Equation Reported cell potentials are typically measured under standard conditions:     Solutes in aqueo...
 It has been determined that cell potentials are related to concentrations of reactants  and products, and to temperature...
II   Using the Nernst Equation to Illustrate the Effect of Dilution on (E— )cell                                          ...
Scenario 1: How does (E— )cell change when water is added to the Ni2+/Ni half cell such that [Ni2+]               o  is d...
Scenario 2: How does (E— )cell change when water is added to the Al3+/Al half cell such that [Al3+]               o  is d...
Scenario 3: How does (E— )cell change when water is added to both half cells such that [Al3+] and                o  [Ni2+...
ConclusionIn general, any change to the cell that increases Q decreases Ecell, while any changethat decreases Q will incre...
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Effect of Concentration Changes on Cell Potential

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[ Visit http://www.wewwchemistry.com ] This example uses the Nernst equation to illustrate how changes in reactant or product concentration (effected by dilution) affect cell potentials.

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Effect of Concentration Changes on Cell Potential

  1. 1. 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— )cell, the cell potential measured under o standard conditions.http://www.wewwchemistry.com Page 1
  2. 2.  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— )cell – ⎜ RT ⎟ ln Q o ⎛ ⎞ ⎜ nF ⎟ ⎝ ⎠ 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: Ecell = (E— )cell – ⎜ 0.0592 ⎟ log10 Q o ⎛ ⎞ ⎜ ⎝ n ⎟⎠http://www.wewwchemistry.com Page 2 €
  3. 3. II Using the Nernst Equation to Illustrate the Effect of Dilution on (E— )cell o 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— )cell = (E— )reduction – (E— )oxidation o o o = –0.25 – (–1.66) = +1.41 Vhttp://www.wewwchemistry.com Page 3
  4. 4. Scenario 1: How does (E— )cell change when water is added to the Ni2+/Ni half cell such that [Ni2+] o is decreased ten times? (All measurements are taken at 298 K.) — )cell – ⎛ 0.0592 ⎞ log10 Q o Ecell = (E ⎜ ⎜ n ⎟ ⎟ ⎝ ⎠ 3+ 2 +1.41 – ⎜ 0.0592 ⎟ log10 [Al ] ⎛ ⎞ = 6 ⎟ € ⎜ ⎝ ⎠ [Ni2+ ]3 +1.41 – ⎜ 0.0592 ⎟ log10 1 ⎛ ⎞ = ⎜ 6 ⎟ ⎞3 ⎜ 1 ⎟ ⎝ ⎠ ⎛ € € ⎜ 10 ⎟ ⎝ ⎠ = +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.http://www.wewwchemistry.com Page 4
  5. 5. Scenario 2: How does (E— )cell change when water is added to the Al3+/Al half cell such that [Al3+] o is decreased ten times? (All measurements are taken at 298 K.) — )cell – ⎛ 0.0592 ⎞ log10 Q o Ecell = (E ⎜ ⎜ n ⎟ ⎟ ⎝ ⎠ 3+ 2 +1.41 – ⎜ 0.0592 ⎟ log10 [Al ] ⎛ ⎞ = 6 ⎟ € ⎜ ⎝ ⎠ [Ni2+ ]3 2 ⎛ ⎜ 1 ⎞ ⎟ = € 0.0592 ⎞ € ⎛ ⎜ 10 ⎟ +1.41 – ⎜ ⎟ log10 ⎝ ⎠ ⎜ ⎝6 ⎟⎠ 1 = +1.41 – (–0.0197) = +1.43 V € € (Q < 1) Ecell increases.http://www.wewwchemistry.com Page 5
  6. 6. Scenario 3: How does (E— )cell change when water is added to both half cells such that [Al3+] and o [Ni2+] are each decreased by ten times? (All measurements are taken at 298 K.) — )cell – ⎛ 0.0592 ⎞ log10 Q o Ecell = (E ⎜ ⎜ n ⎟ ⎟ ⎝ ⎠ 3+ 2 +1.41 – ⎜ 0.0592 ⎟ log10 [Al ] ⎛ ⎞ = 6 ⎟ € ⎜ ⎝ ⎠ [Ni2+ ]3 2 ⎛ ⎜ 1 ⎞ ⎟ € 0.0592 ⎞ € ⎛ ⎜ 10 ⎟ = +1.41 – ⎟ log10 ⎜ ⎝ ⎠ 6 ⎟ ⎜ ⎝ ⎠ ⎛ 1 ⎞ 3 ⎜ ⎟ ⎜ 10 ⎟ ⎝ ⎠ = +1.41 – 0.00987 € = +1.40 V € (Resulting Q > 1) Ecell decreases, albeit slightly.http://www.wewwchemistry.com Page 6
  7. 7. ConclusionIn general, any change to the cell that increases Q decreases Ecell, while any changethat decreases Q will increase Ecell. Thus, adding reactant or removing productincreases 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— )cell (Scenario 1). o when product concentration ([Al3+]) is smaller than 1 mol dm–3, Ecell is more than (E— )cell (Scenario 2). o If concentrations of both products and reactants are decreased in the cell, whether Ecell is more or less than (E— )cell depends on the resulting Q: o  Q > 1, Ecell is less than (E— )cell (Scenario 3) o  Q = 1, there is no change to (E— )cell, i.e. Ecell = (E— )cell o o  Q < 1, Ecell is more than (E— )cell ohttp://www.wewwchemistry.com Page 7

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