Nernst Cell Voltage Equation
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Nernst Cell Voltage Equation

Given the following Daniell cell (a typical voltaic or galvanic cell):

Zn|Zn2+||Cu2+|Cu

Determine the cell voltage (potential) at 25oC If the concentrations of the zinc ions and the copper ions are 0.50m and 0.20m respectively.

The strings:

S7P5A52 (Change - Chemical Change)

The math:
Pj Problem of Interest is of type change (chemical change). This problem is also a force problem because we are determining voltage which is force (push). However, the Pj Problem of Interest is cast as of type change because of the underlying chemical reaction that made the cell voltage possible.

This problem is solved using Nernst equation:

Cell voltage, E = Eo - (RT/nF)lnK
where K = equilibrium constant = concentration of products/concentration of reactants.

The half-reactions of the cell are as follows:

Half-reaction at the anode:

Zn -------> Zn2+ + 2e- --------(1) oxidation reaction.
Standard reduction potential from table = - 0.762 volts
Since this half-reaction is an oxidation reaction, we change -0.762 to +0.762.

Half-reaction at the cathode:

Cu2+ + 2e- -------> Cu -------(2) reduction reaction
Standard reduction potential from table = 0.340
Since this half-reaction is a reduction reaction, we do not need to change table value.

Overall cell reaction

Zn(c) + Cu2+(aq) --------> Zn2+(aq) + Cu(c)---------(3)
So, Eo = 0.762 + 0.340 = 1.102 volts.
Since concentrations are not standard (1M), we use the Nernst equation.

So, E = 1.102 - (8.31)(298)/2(96485)ln(0.50/0.20)
So, cell voltage, E = 1.102 - (0.01284)ln2.5 = 1.088 volts.

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