Cell Potential - Chemistry

1. Fundamental Concepts

Definition: Cell potential is the measure of the electric potential difference between two half-cells in an electrochemical cell.
Standard Reduction Potential: The tendency of a chemical species to be reduced under standard conditions, measured in volts (V).
Nernst Equation: Relates the cell potential to the concentrations of the reactants and products in the cell.

2. Key Concepts

Standard Cell Potential: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

Nernst Equation: $E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q$

Application: Used to predict the spontaneity of redox reactions and to calculate the cell potential under non-standard conditions.

3. Examples

Example 1 (Basic)

Problem: Calculate the standard cell potential for the following reaction: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$ Given: $$ E_{Cu^{2+}/Cu}^{\circ} = 0.34 V $$, $$ E_{Zn^{2+}/Zn}^{\circ} = -0.76 V $$

Step-by-Step Solution:

Identify the cathode and anode: Cathode: $$ Cu^{2+}/Cu $$, Anode: $$ Zn^{2+}/Zn $$
Use the standard reduction potentials: $$ E_{cathode}^{\circ} = 0.34 V $$, $$ E_{anode}^{\circ} = -0.76 V $$
Calculate the standard cell potential: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.34 V - (-0.76 V) = 1.10 V$

Validation: The calculated cell potential is positive, indicating a spontaneous reaction.

Example 2 (Intermediate)

Problem: Calculate the cell potential for the following reaction at 25°C: $Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$ Given: $$ E_{Ag^+/Ag}^{\circ} = 0.80 V $$, $$ E_{Fe^{3+}/Fe^{2+}}^{\circ} = 0.77 V $$, $$ [Fe^{2+}] = 0.1 M $$, $$ [Fe^{3+}] = 0.01 M $$, $$ [Ag^+] = 0.01 M $$

Step-by-Step Solution:

Identify the cathode and anode: Cathode: $$ Ag^+/Ag $$, Anode: $$ Fe^{3+}/Fe^{2+} $$
Use the standard reduction potentials: $$ E_{cathode}^{\circ} = 0.80 V $$, $$ E_{anode}^{\circ} = 0.77 V $$
Calculate the standard cell potential: $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 0.80 V - 0.77 V = 0.03 V$
Use the Nernst equation to find the cell potential at non-standard conditions: $E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln Q$ where $$ Q = \frac{[Fe^{3+}]}{[Fe^{2+}][Ag^+]} = \frac{0.01}{0.1 \times 0.01} = 1 $$
Substitute the values into the Nernst equation: $E_{cell} = 0.03 V - \frac{(8.314 J/(mol \cdot K) \times 298 K)}{(1 mol \times 96485 C/mol)} \ln 1 = 0.03 V - 0 = 0.03 V$

Validation: The calculated cell potential is 0.03 V, which is consistent with the standard cell potential since the reaction quotient $$ Q $$ is 1.

4. Problem-Solving Techniques

Identify Half-Reactions: Determine the oxidation and reduction half-reactions and their standard reduction potentials.
Use Standard Reduction Potentials: Use the standard reduction potentials to calculate the standard cell potential.
Apply the Nernst Equation: Use the Nernst equation to calculate the cell potential under non-standard conditions.
Check Spontaneity: Ensure the cell potential is positive to confirm the reaction is spontaneous.