What is the value of the equilibrium constant at 25°C for the reaction between Al(s) and Cd^2+(aq)?

Saturday, April 21, 2012

What is the value of the equilibrium constant at 25°C for the reaction between Al(s) and Cd^2+(aq)?

The half cell reactions and the standard electrode potentials at 25 degrees Celsius for the given species are:

$\displaystyle{A}{{l}}^{{{3}+}}+{3}{{e}}^{{-}}\to{A}{l}{\left({s}\right)}$

$\displaystyle{C}{{d}}^{{{2}+}}+{2}{{e}}^{{-}}\to{C}{d}{\left({s}\right)}$

Standard electrode potentials are -1.66 V and -0.4 V

The overall cell reaction between Cd2+ (aq) and Al (s) can be written as:

$\displaystyle{A}{l}{\left({s}\right)}+{C}{{d}}^{{{2}+}}\to{A}{{l}}^{{{3}+}}{\left({a}{q}\right)}+{C}{d}{\left({s}\right)}$

and the balanced cell reaction can be written as:

$\displaystyle{2}{A}{l}{\left({s}\right)}+{3}{C}{{d}}^{{{2}+}}{\left({a}{q}\right)}\to{2}{A}{{l}}^{{{3}+}}{\left({a}{q}\right)}+{3}{C}{d}{\left({s}\right)}$

At equilibrium, standard cell potential, E' = +1.66 + (-0.4) V = 1.26 V

Using Nernst equation: E' = (RT/nF) ln K

where, R = 8.314 J/k/mol, n = 6, F = 96500 C/mol, T = 298 K

we get: 1.26 = (8.314 x 298)/(6 x 96500) ln K

solving the equation, we get, K = 7.6 x 10^127.

Thus, the equilibrium constant at 25 degrees C for the given cell is 7.3 x 10^127.

Hope this helps.

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Saturday, April 21, 2012