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Nernst Equation

Electrochemical cell problems – Using Nernst equation solve the following problems

First watch the video and then attempt to solve the problem

An electrochemical cell is set up using Aluminum and Copper electrodes in their respective solutions. If the concentration of aluminum ions is 0.0001M and the concentration of copper ions are 0.001M, determine the cell potential of the resulting electrochemical cell. Standard reduction potential of Al^{3+}|Al is -1.66 V and Cu^{2+}|Cu is 0.34V

For the following electrochemical cell determine the emf of the cell using the Nernst equation Mg(s) | Mg^{2+}(0.001M) ||`Cu^{2+}(0.01M) | Cu(s). Standard reduction potential of Mg is taken as -2.38 V and Cu is 0.34 V.

Steps to solve Nernst equation

Decide which electrode is the anode and cathode from standard reduction potential values

The electrode with larger reduction potential value will under go reduction or will be the cathode, the electrode with a smaller value will be the anode

At the anode the reaction is oxidation, you can write the OHR or oxidation half reaction

At the cathode the reaction is reduction, you can write the RHR or reduction half reaction

Once you have the two half reactions you can write the “Net cell reaction” or the ” Net ionic equation” for the reaction from which the value of “n” can be determined.

‘n’ represents the number of electrons transacted in the balance equation

From the net ionic equation write the Kc expression, if a negative sign is used in the Nernst equation Kc is the ratio of the concentration of [Anion]/[Cation] raised to the appropriate coefficients from the balanced equation.

Calculate E^{o}Cell = E^{o}cathode – E^{o}anode

Now substitute all the values in the Nernst equation

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