Omar asked in Science & MathematicsChemistry · 2 months ago


In the concentration cell shown, Ecell = +0.0296 V at 298 K. What is the concentration at the cathode?

The concentration of the anode is .01M Fe2+


In the electrochemical cell shown in the question above, what concentration of Fe2+ at the cathode would be required to reach a cell potential of 0.05 V at 298 K?

Any help greatly appreciated

1 Answer

  • 2 months ago

    The voltage of a concentration cell is given as:

    E = -0.0592/n log [Fe2+]anode/[Fe2+]cathode

    0.0296 V = -0.0592 / 2 log 0.01/[Fe2+]

    [Fe2+] = 0.10 M

    Use the same equation,

    0.050 V = -0.0592/2 log (0.01/[Fe2+]

    [Fe2+] = 0.50 M

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