Math Problem Statement

Ni-Cad (nickel-cadmium) batteries have a slightly lower cell potential than the common alkaline value of 1.5 V normally used in flashlights. They have a cadmium anode, which is oxidized to Cd(OH)2(s) in a basic background electrolyte of concentrated OH-, and a cathode half-reaction in which NiO(OH)(s) is reduced to Ni(OH)2(s). What is the value of ΔG°cell?

Solution

To determine the Gibbs free energy change, ΔGcell\Delta G^\circ_{\text{cell}}, for the given Ni-Cad battery reaction, we can use the following formula:

ΔGcell=nFEcell\Delta G^\circ_{\text{cell}} = -n F E^\circ_{\text{cell}}

where:

  • nn is the number of moles of electrons transferred,
  • FF is Faraday's constant (96485C/mol e96485 \, \text{C/mol e}^-),
  • EcellE^\circ_{\text{cell}} is the standard cell potential.

Step 1: Calculate EcellE^\circ_{\text{cell}}

The cell potential EcellE^\circ_{\text{cell}} can be calculated by finding the difference between the reduction potentials of the cathode and the anode.

The reduction half-reactions and their standard reduction potentials are:

  1. Cathode (Reduction): NiO(OH)(s)+2H2O(l)+2eNi(OH)2(s)+2OH(aq)\text{NiO(OH)}(s) + 2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{Ni(OH)}_2(s) + 2\text{OH}^-(aq) with Ered=1.32VE^\circ_{\text{red}} = 1.32 \, V
  2. Anode (Oxidation): Cd(OH)2(s)+2eCd(s)+2OH(aq)\text{Cd(OH)}_2(s) + 2e^- \rightarrow \text{Cd}(s) + 2\text{OH}^-(aq) with Ered=0.403VE^\circ_{\text{red}} = -0.403 \, V

To find EcellE^\circ_{\text{cell}}: Ecell=EcathodeEanode=1.32V(0.403V)=1.32V+0.403V=1.723VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.32 \, V - (-0.403 \, V) = 1.32 \, V + 0.403 \, V = 1.723 \, V

Step 2: Determine nn, the number of electrons

From the balanced half-reactions, 2 moles of electrons are transferred in the reaction.

Step 3: Calculate ΔGcell\Delta G^\circ_{\text{cell}}

Now, substitute the values into the Gibbs free energy formula: ΔGcell=nFEcell\Delta G^\circ_{\text{cell}} = -n F E^\circ_{\text{cell}} ΔGcell=(2)(96485C/mol e)(1.723V)\Delta G^\circ_{\text{cell}} = - (2) (96485 \, \text{C/mol e}^-) (1.723 \, V)

Calculating this: ΔGcell=2×96485×1.723=332,728.61J=332.73kJ\Delta G^\circ_{\text{cell}} = -2 \times 96485 \times 1.723 = -332,728.61 \, \text{J} = -332.73 \, \text{kJ}

Final Answer:

ΔGcell=332.73kJ\Delta G^\circ_{\text{cell}} = -332.73 \, \text{kJ}

Would you like further clarification on any step, or do you have additional questions?

Here are 5 related questions to explore:

  1. How does the standard reduction potential relate to the spontaneity of a reaction?
  2. What happens to ΔG\Delta G^\circ if the cell potential is reversed?
  3. Why is Faraday's constant important in electrochemical calculations?
  4. How would the cell potential change if the concentration of OHOH^- ions varied?
  5. What is the impact of temperature on the cell potential?

Tip: Always verify the number of electrons transferred when calculating Gibbs free energy changes in electrochemical reactions, as it can vary between reactions.

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Math Problem Analysis

Mathematical Concepts

Electrochemistry
Gibbs Free Energy
Cell Potential

Formulas

ΔG°cell = -nFE°cell
E°cell = E°cathode - E°anode

Theorems

Gibbs Free Energy Change for Electrochemical Cells

Suitable Grade Level

College (Chemistry)