go to the AUS-e-TUTE homepage

Nernst Equation Tutorial

Key Concepts

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

Worked Example of E Calculation

Question:

Calculate the voltage produced by the cell Sn(s)|Sn2+||Ag+|Ag(s) at 25°C given:
[Sn2+] = 0.15 mol L-1
[Ag+] = 1.7 mol L-1

Solution:

  1. Write the Nernst Equation for 25°C:
    E = Eo -0.0592/n × log10Q
  2. Calculate Eo for the cell:
    anode: Sn(s) Sn2+ + 2e Eo = +0.14 V
    cathode: 2 × [e- + Ag+ Ag(s)] Eo = +0.80 V

    cell: Sn(s) + 2Ag+ Sn2+ + 2Ag(s) Eo = +0.94 V
  3. Wite the expression for Q:
    Q = [Sn2+]/[Ag+]2
    (concentrations of solids is constant and incorporated in the value of Q)
  4. Write the Nernst Equation for this example:
    E = Eo -0.0592/n × log10([Sn2+]/[Ag+]2)
  5. Substitute the values:
    Eo = +0.94 V
    n = 2 (2 moles of electrons transferred during the redox reaction)
    [Sn2+] = 0.15 mol L-1
    [Ag+] = 1.7 mol L-1

    E = +0.94 -0.0592/2 × log([0.15]/[1.7]2)

  6. Calculate Q:
    E = +0.94 -0.0592/2 × log10[0.0519]
  7. Calculate logQ:
    E = +0.94 -0.0592/2 × -1.285
  8. Calculate E:
    E = +0.94 -0.0592/2 × -1.285
    E = +0.98 V

Note: E > Eo, and positive, so the cell reaction has a greater tendency to take place at these concentrations.

Do you know this?

Join AUS-e-TUTE!

Play the game now!

Worked Example of K Calculation

Question:

Calculate the equilibrium constant, K, for the reaction Sn(s)|Sn2+||Ag+|Ag(s) at 25°C.

Solution:

  1. Write the Nernst Equation:

    E = Eo -(0.0592/n)log10Q

  2. Write the Nernst Equation for the equilibrium expression:
    At equilibrium Q = K and E = 0

    Eo = (0.0592/n)log10K

  3. Calculate Eo for the equation (as above) = +0.94 V
  4. Calculate n (moles of electrons transferred) = 2
  5. Substitute values for Eo and n into the equation:
    +0.94 = (0.0592/2)log10K
  6. Calculate log10K:
    +0.94 = 0.0296log10K
    log10K = 31.76
  7. Calculate K:
    K = 1031.76 = 5.71 × 1031

Note: K is very large so the forward reaction is favoured and the equilibrium position lies very far to the right.

Do you understand this?

Join AUS-e-TUTE!

Take the test now!