# 23.7: Laboratory Activities for Chapter 23

**At Grade**Created by: CK-12

## Dependence of Cell Potential on Concentration

*Non-Standard Cells*

For the most part, the discussion of cell potential in high school chemistry deals with cells under standard conditions. Standard conditions for cells is \begin{align*}25^ \circ \ C, 1.0 \ atm\end{align*} pressure, and the concentrations of ions is \begin{align*}1.0 \ M\end{align*}. If you were building a cell to use to do work, you would not build a standard cell. Standard cells are used essentially for teaching or experimentation. They do not have either the maximum voltage, or the maximum capacity, that can be built into Galvanic cells. The advantage of standard cells is that their voltages are precisely predictable and easily calculated.

The cell potential for galvanic cells is closely related to the net movement of materials from reactants to products. A faster net reaction would produce a greater cell potential, and a slower net reaction would produce a smaller cell potential. At equilibrium, the net reaction is zero and therefore, the cell potential would be zero. If the forward reaction rate is increased with no change in the reverse reaction rate, then the net forward reaction is greater, and the cell potential will be greater. If the reverse reaction rate is decreased with no chain in the forward rate, then the net forward reaction is greater, and the cell potential will be greater.

Consider the cell composed of the standard half cells of aluminum and manganese.

\begin{align*}2Al_{(s)} \ + \ 3Mn^{2+}_{(aq)} \ \rightarrow \ 2 \ Al^{3+}_{(aq)} \ + \ 3 Mn_{(s)} \qquad \ E^ \circ _{cell} \ = \ 0.48 \ V\end{align*}

The \begin{align*}E^ \circ _{cell}\end{align*} for this reaction is determined when the concentrations of manganese ion and the aluminum ion are both \begin{align*}1.0 \ M\end{align*}. This is the voltage of this cell at standard conditions. If the concentration of the manganese ion is increases, the forward reaction rate will increase, and the net movement of material in the forward direction will increase. This increase in the net movement of material in the forward direction will cause the cell voltage to be higher. If the concentration of the aluminum ion is decreased, the reverse reaction rate will decrease, and the net movement of material in the forward direction will increase. This increase in the net movement of material in the forward direction causes the voltage to be higher. If the \begin{align*}[Mn^{2+}]\end{align*} concentration is decreased, the forward reaction rate will decrease, and the net movement of material in the forward direction will decrease. Therefore, the voltage of the cell will be lower. If the \begin{align*}[Al^{3+}]\end{align*} concentration is increased, the reverse reaction rate will be increased, and the net movement of material in the forward direction will decrease. Therefore, the voltage of the cell will be lower.

Cells that do not have the concentrations of ions at \begin{align*}1.0 \ M\end{align*} are called non-standard cells. In the cell described above, \begin{align*}[Mn^{2+}] \ > \ 1.0 \ M\end{align*} will cause the cell voltage to be greater than \begin{align*}0.48 \ V\end{align*} and \begin{align*}[Al^{3+}] \ > \ 1.0 \ M\end{align*} will cause the cell voltage to be less than \begin{align*}0.48 \ volts\end{align*}. Vice versa would be true if the concentrations were less than \begin{align*}1.0 \ M\end{align*}.

Cell voltages for non-standard cells can also be calculated using the Nernst Equation.

The Nernst Equation is \begin{align*}E = E^ \circ - \left ( \frac{0.0591}{n} \right )(log Q)\end{align*}, where \begin{align*}E\end{align*} is the voltage of the non-standard cell. \begin{align*}E ^ \circ \end{align*} would be the voltage of these reactants and products if they were a standard cell, \begin{align*}n\end{align*} is the moles of electrons transferred in the balanced reaction, and \begin{align*}Q\end{align*} is the reaction quotient. The reaction quotient is the equilibrium constant expression, but when the reaction is not at equilibrium it is called the reaction quotient.

The reaction quotient for the example cell used here is \begin{align*}Q = \frac {[Al^{3+}]^2}{[Mn^{2+}]^3} \end{align*}.

The Nernst Equation for this cell is \begin{align*}E = E^ \circ - \left ( \frac{0.0591}{n} \right ) \left (log \frac {[Al^{3+}]^2} {[Mn^{2+}]^3} \right )\end{align*}.

If you follow the mathematics for the case when both ions concentrations are \begin{align*}1.0 \ M\end{align*}, the reaction quotient would be \begin{align*}1\end{align*} and the \begin{align*}\log\end{align*} of \begin{align*}1\end{align*} is zero. Therefore, the second term in the Nernst Equation is zero and \begin{align*}E \ = \ E^ \circ \end{align*}.

Let’s take the case of the example cell when \begin{align*}[Mn^{2+}] \ = \ 6.0 \ M\end{align*} and \begin{align*}[Al^{3+}] \ = \ 0.10 \ M\end{align*}.

\begin{align*}E &= E^ \circ - \left (\frac {0.0591}{n} \right ) \left ( log \frac{[Al^{3+}]^2}{[Mn^{2+}]^3} \right )\\ E &= 0.48 \ V - \left ( \frac {0.0591}{6} \right ) \left ( log \frac{[0.10]^2}{[6.0]^3} \right )\\ E &= 0.48 \ V - \left ( \frac{0.0591}{6} \right )(-4.33)\\ E &= 0.48 \ V - (-0.04 \ V)\\ E &= 0.52 \ V\end{align*}

*Concentration Cells*

If we attempt to construct a standard cell from the same two reactants, we do not get a reaction or a cell voltage. Suppose we attempt to build a cell with two silver half-cells.

\begin{align*}Ag_{(s)} \ + \ Ag^+_{(aq)} \ \rightarrow \ Ag^+_{(aq)} \ + \ Ag_{(s)}\end{align*}

If this is a standard cell, the half-cell voltage for the oxidation half-reaction is \begin{align*}-0.80 \ V\end{align*} and the half-cell voltage for the reduction half-reaction is \begin{align*}+0.80 \ V\end{align*}. Clearly the net voltage is . It is possible, however, to produce a voltage using the same two half-reactions if we alter the concentrations of the ions. Such a cell is called a concentration cell and its voltage can be calculated using the Nernst Equation.

The Nernst Equation would look like this:

\begin{align*}E = E^ \circ - \left ( \frac{0.0591}{n} \right ) \left ( log \frac {[Ag^+]}{[Ag^+]} \right )\end{align*}, where the silver ion concentration in the numerator is the concentration of the silver ion in the products and the silver ion concentration in the denominator is the silver ion concentration in the reactants.

Suppose we build this cell using a concentration of silver ion in the products of \begin{align*}0.010 \ M\end{align*} and a silver ion concentration in the reactants of \begin{align*}6.0 \ M\end{align*}. The \begin{align*}E^ \circ\end{align*} in this case is zero and \begin{align*}n = 1\end{align*}.

\begin{align*}E &= E^ \circ - \left ( \frac {0.0591} {n} \right ) \left ( log \frac {[Ag^+]} {[Ag^+]} \right )\\ E &= 0 \ V - (0.0591) \left ( log \frac {[0.010]}{[6.0]} \right )\\ E &= 0 \ V - (0.0591)(- 2.78 \ V)\\ E &= 0 \ V - (- 0.16 \ V)\\ E &= 0.16 \ V\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Show More |