Electrochemical Cell EMF Example Problem

The cell electromotive force, or cell EMF, is the net voltage between the oxidation and reduction half-reactions taking place between two redox half-reactions. Cell EMF is used to determine whether or not the cell is galvanic. This example problem shows how to calculate the cell EMF using standard reduction potentials.
The Table of Standard Reduction Potentials is needed for this example. In a homework problem, you should be given these values or else access to the table.

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Equilibrium Constant of an Electrochemical Cell

By Todd Helmenstine

Sample EMF Calculation

Consider the redox reaction:

• Mg(s) + 2 H⁺(aq) → Mg²⁺(aq) + H₂(g)
  • a) Calculate the cell EMF for the reaction.
  • b) Identify if the reaction is galvanic.
• Solution:
  • Step 1: Break the redox reaction into reduction and oxidation half-reactions.
    Hydrogen ions, H⁺ gain electrons when forming hydrogen gas, H₂. The hydrogen atoms are reduced by the half-reaction:
    2 H⁺ + 2 e^- → H₂
    Magnesium loses two electrons and is oxidized by the half-reaction:
    Mg → Mg²⁺ + 2 e^-
  • Step 2: Find the standard reduction potentials for the half-reactions.
    Reduction: E⁰ = 0.0000 V
    The table shows reduction half-reactions and standard reduction potentials. To find E⁰ for an oxidation reaction, reverse the reaction.
  • Reversed reaction:
    Mg²⁺ + 2 e^- → Mg
    This reaction has a E₀ = -2.372 V.
    E⁰_Oxidation = - E⁰_Reduction
    E_0Oxidation = - (-2.372 V) = + 2.372 V
  • Step 3: Add the two E⁰ together to find the total cell EMF, E⁰_cell
    E⁰_cell = E⁰_reduction + E⁰_oxidation
    E⁰_cell = 0.0000 V + 2.372 V = +2.372 V
  • Step 4: Determine if the reaction is galvanic. Redox reactions with a positive E⁰_cell value are galvanic.
    This reaction's E⁰_cell is positive and therefore galvanic.
• Answer:
  The cell EMF of the reaction is +2.372 Volts and is galvanic.