Nernst Equation

1
Nernst Equation

• If E cell is positive, the reaction in the
  forward direction (from left to right) is
  spontaneous.
• If E cell is negative, the reaction in the
  forward direction is nonspontaneous
• If E cell 0, the system is at equilibrium
• When a cell reaction is reversed, E cell change
  signs.

VIDEO 1
2
Nernst Equationelectrochemical cells under
non-standard conditions

• Ecell E0cell - (2.303RT/nF) log Q
• R 8.31 JK-1 mol-1
• n number of electrons transferred
• F Faraday constant 96,500 JV-1 mol-1
• Q reaction quotient
• Ecell E0cell - (0.0592/n) log Q

VIDEO 2
VIDEO 3
3

• Relates a cell voltage for nonstandard
  conditions, E cell, to a standard cell voltage, E
  cell0, and the concentrations of reactants and
  products expressed through the reaction quotient,
  Q.
• In the equation, n is the number of moles of
  electrons involved in the overall cell reaction.

4
Example 1

• Will copper metal displace silver ion from
  aqueous solution? That is, does this reaction
  occur spontaneously from left to right?
• Cu(s) 2Ag ? Cu2 2 Ag

5
Solution

• Reduction 2 Ag e ? Ag
• Oxidation Cu ? Cu2 2e
• Overall Cu 2Ag ? Cu2 2 Ag

6

• E0 cell E0 (cathode) - E0 (anode)
• E0 Ag/Ag - E0 Cu2/Cu
• 0.800 V 0.340 V
• 0.460 V
• Because E0 cell is positive, the forward
  direction should be the direction of spontaneous
  change.
• Copper metal should displace silver ions from
  solution.

7
Example

• Calculate the expected voltmeter reading for the
  voltaic cell pictured in figure below.

e-
Platinum electrode
Cu2 0.5 M
Fe2 0.1M
Salt bridge
Copper electrode
Fe3 0.2M
8
Solution

• The direction of electron flow enables us to
  identify the anode and cathode.
• Anode(ox) Cu ? Cu2 (0.5M) 2e-
• Cathode(red) 2 Fe3 (0.5M) e- ? Fe2 (0.1M)
• Overall 2Fe3 (0.5M) Cu(s) ? Cu2 (0.5M)
  2Fe2(0.1M)

9
Solution

• 1. Calculate E0 cell using the value from table.
• E0 cell E0 (reduction) E0 (oxidation)
• E0 Fe3/Fe2 E0 Cu/Cu2
• 0.771 V ( 0.340 V)
• 0.431 V

10

• 2. Using Nernst eq
• E cell E0 cell - 0.0592 log Q
• n
• E0 cell - 0.0592 log products
• n reactants
• 0.431 0.0592 log Cu2Fe22
• 2 Fe32
• 0.431 0.0592 log
  (0.5)(0.1)2
• 2 (0.2)2
• 0.458 V

VIDEO 4
11
Exercise Below is the cell notation of an
electrochemical cell.

• Mg(s) l Mg2(0.5M) ll Fe3 (0.1M),Fe2(0.5M) l
  Pt(s)
• Given Eo Fe3/Fe2 0.77 V
• Eo Mg2/Mg -2.38 V
• Calculate the electrode potential, E cell.

12
Solution

• Anode(ox) Mg (s)? Mg2 (0.5M) 2e-
  E0 2.38V
• Cathode(red) 2 Fe3 (0.1M) e- ? Fe2 (0.5M)
  E0 0.77V
• Overall 2Fe3 (1M) Mg(s) ? Mg2 (0.5M)
  2Fe2(0.5M)

13

• E0 cell E0 red E0 ox
• E0 Fe3/Fe2 E0 Mg/Mg2
• (0.77 2.38) V
• 3.15 V

VIDEO 5
14

• Using Nernst eq
• E cell E0 cell - 0.0592 log Q
• n
• 3.15 0.0592 log Mg2Fe22
• 2 Fe32
• 3.15 0.0592 log (0.5)(0.5)2
• 2 (0.1)2
• 3.118 V

15
Example
Calculate the equilibrium constant (K) for the
following reaction.
Cu(s) 2Ag(ak) Cu2(ak)
2Ag(s)
Answer
At equilibrium, E cell 0
Eocell Eo cathode - Eo anode
0.80 ( 0.34) 0.46 V
16
Ecell Eocell 0.0592 log K
2
0 0.46 0.0592 log K
2
log K 15.54 K 3.467 x 1015
VIDEO 6
17
Example

• The cell potential measured for the
  electrochemical cell represented by the cell
  diagram below is 0.70V at 25C. Calculate the
  concentration of H ions used in the cell.
• Zn(s) Zn2(aq,1 moldm-3) H(aq, X
  moldm-3H2(1atm)Pt(s)

VIDEO 7