# 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*}

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*}

Cells that do not have the concentrations of ions at \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*}

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*}

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