Galvanic Cell Potential Calculation
Calculate the standard cell potential for a galvanic cell and determine if the reaction is spontaneous. Master electrochemistry fundamentals.
Problem Scenario
A galvanic cell is constructed using a zinc electrode in 1.0 M Zn(NO3)2 solution and a copper electrode in 1.0 M Cu(NO3)2 solution. Calculate the standard cell potential and write the cell notation.
Given Data
Requirements
- Identify oxidation and reduction half-reactions
- Find standard reduction potentials
- Calculate standard cell potential
- Write cell notation
- Determine if reaction is spontaneous
Solution
Step 1:
Identify half-reactions: Since Zn has the more negative reduction potential, it will be oxidized (anode). Oxidation: Zn(s) -> Zn2+(aq) + 2e-. Reduction: Cu2+(aq) + 2e- -> Cu(s).
Step 2:
Calculate E°cell using the formula: E°cell = E°cathode - E°anode = E°Cu2+/Cu - E°Zn2+/Zn = (+0.34 V) - (-0.76 V) = +1.10 V
Step 3:
Write the overall cell reaction by adding half-reactions: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
Step 4:
Write cell notation (anode on left, cathode on right): Zn(s) | Zn2+(1.0 M) || Cu2+(1.0 M) | Cu(s). Single lines separate phases, double line represents the salt bridge.
Step 5:
Determine spontaneity: Since E°cell = +1.10 V > 0, the reaction is spontaneous under standard conditions. Also, deltaG° = -nFE° = -(2)(96485)(1.10) = -212 kJ/mol < 0.
Final Answer
E°cell = +1.10 V. Cell notation: Zn(s) | Zn2+(1.0 M) || Cu2+(1.0 M) | Cu(s). The positive cell potential indicates this is a spontaneous galvanic cell. This is the basis of common batteries and explains why zinc is more reactive than copper.
Key Takeaways
- ✓E°cell = E°cathode - E°anode (always subtract anode from cathode)
- ✓The metal with more negative E° is oxidized (anode)
- ✓Positive E°cell means spontaneous reaction
- ✓In cell notation, anode is written on the left
Common Errors to Avoid
- ✗Subtracting in the wrong order (cathode - anode, not anode - cathode)
- ✗Changing signs of standard potentials unnecessarily
- ✗Confusing anode and cathode
- ✗Wrong cell notation format
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Common questions about this problem type
In a galvanic cell, the anode is negative because oxidation releases electrons there. Zinc, being more easily oxidized, releases electrons that flow through the external circuit to the cathode.
Use the Nernst equation: E = E° - (0.0592/n)log(Q) at 25°C. Q is the reaction quotient calculated from concentrations of products over reactants.