Nernst Cell Voltage Equation

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{5}A_{52} Base Sequence = 12735 String Sequence = 12735 - 5 - 52 **

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Nernst Cell Voltage Equation

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

Zn|Zn^{2+}||Cu^{2+}|Cu

Determine the cell voltage (potential) at 25^{o}C If the concentrations of the zinc ions and the copper ions are 0.50m and 0.20m respectively.

**The strings**:

S_{7}P_{5}A_{52} (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 = E^{o} - (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 -------> Zn^{2+} + 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:

Cu^{2+} + 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) + Cu^{2+}(aq) --------> Zn^{2+}(aq) + Cu(c)---------(3)

So, E^{o} = 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.

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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