Galvanic (Voltaic) Cells - Chemistry

1. Fundamental Concepts

Definition: A Galvanic (Voltaic) cell is a device that converts chemical energy into electrical energy through spontaneous redox reactions.
Components: Anode (oxidation), Cathode (reduction), Salt bridge, and External circuit.
Cell Potential: The difference in electric potential between the anode and cathode, measured in volts (V).

2. Key Concepts

Redox Reactions: $$\text{Oxidation: } \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$$ $$\text{Reduction: } \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$$

Standard Cell Potential: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$

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

3. Examples

Example 1 (Basic)

Problem: Calculate the standard cell potential for a galvanic cell with the following half-reactions: $$\text{Anode: } \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad E^\circ = -0.76 \, \text{V}$$ $$\text{Cathode: } \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = +0.34 \, \text{V}$$

Step-by-Step Solution:

Identify the standard reduction potentials: $$ E^\circ_{\text{anode}} = -0.76 \, \text{V} $$ and $$ E^\circ_{\text{cathode}} = +0.34 \, \text{V} $$.
Use the formula for standard cell potential: $$ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} $$.
Substitute the values: $$ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 1.10 \, \text{V} $$.

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

Example 2 (Intermediate)

Problem: Calculate the cell potential at 25°C for a galvanic cell with the following concentrations: $$[\text{Zn}^{2+}] = 0.1 \, \text{M}, [\text{Cu}^{2+}] = 0.01 \, \text{M}$$ Given: $$ E^\circ_{\text{cell}} = 1.10 \, \text{V} $$.

Step-by-Step Solution:

Write the Nernst equation: $$ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q $$.
Calculate the reaction quotient $$ Q $$: $$ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} $$.
Substitute the values: $$ Q = \frac{0.1 \, \text{M}}{0.01 \, \text{M}} = 10 $$.
Substitute into the Nernst equation: $$ E_{\text{cell}} = 1.10 \, \text{V} - \frac{(8.314 \, \text{J/mol·K})(298 \, \text{K})}{(2)(96485 \, \text{C/mol})} \ln 10 $$.
Simplify: $$ E_{\text{cell}} = 1.10 \, \text{V} - 0.0295 \, \text{V} \cdot 2.303 \approx 1.04 \, \text{V} $$.

Validation: The calculated cell potential is slightly less than the standard potential, which is expected due to the non-standard conditions.

4. Problem-Solving Techniques

Identify Half-Reactions: Clearly label the anode and cathode half-reactions and their standard reduction potentials.
Use Standard Tables: Refer to standard reduction potential tables to find $$ E^\circ $$ values.
Apply Nernst Equation: Use the Nernst equation to account for non-standard conditions, ensuring correct units and constants.
Check Sign Conventions: Ensure the sign of the cell potential is positive for spontaneous reactions.