Math Problem Statement

Using the following standard reduction potentials.

Fe3+ (aq) + e- → Fe2+ (aq) E° = +0.77 V Pb2+ (aq) + 2 e- → Pb(s) E° = -0.13 V

Calculate the standard cell potential for the galvanic cell reaction given below, and determine whether or not this reaction is spontaneous under standard conditions.

Pb2+ (aq) + 2 Fe2+ (aq) → 2 Fe3+ (aq) + Pb(s)

Solution

To determine the standard cell potential (EcellE^\circ_{\text{cell}}) for the given galvanic cell reaction and whether the reaction is spontaneous under standard conditions, we need to use the standard reduction potentials provided.

Given reduction potentials:

  1. Fe3+(aq)+eFe2+(aq)E=+0.77V\text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq) \quad E^\circ = +0.77 \, \text{V}
  2. Pb2+(aq)+2ePb(s)E=0.13V\text{Pb}^{2+} (aq) + 2 e^- \rightarrow \text{Pb} (s) \quad E^\circ = -0.13 \, \text{V}

The overall reaction is:

Pb2+(aq)+2Fe2+(aq)2Fe3+(aq)+Pb(s)\text{Pb}^{2+} (aq) + 2 \text{Fe}^{2+} (aq) \rightarrow 2 \text{Fe}^{3+} (aq) + \text{Pb} (s)

First, we identify the half-reactions and their standard reduction potentials:

  1. Fe3+(aq)+eFe2+(aq)E=+0.77V\text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq) \quad E^\circ = +0.77 \, \text{V}
  2. Pb2+(aq)+2ePb(s)E=0.13V\text{Pb}^{2+} (aq) + 2 e^- \rightarrow \text{Pb} (s) \quad E^\circ = -0.13 \, \text{V}

For the cell reaction, we need to determine which species is being oxidized and which is being reduced. The overall reaction can be split into its half-reactions, and we reverse one of the reduction reactions to represent oxidation.

Reduction half-reaction (cathode): Fe3+(aq)+eFe2+(aq)Ered=+0.77V\text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq) \quad E^\circ_{\text{red}} = +0.77 \, \text{V}

Oxidation half-reaction (anode) (reversed from the given reduction potential): Pb(s)Pb2+(aq)+2eEox=+0.13V\text{Pb} (s) \rightarrow \text{Pb}^{2+} (aq) + 2 e^- \quad E^\circ_{\text{ox}} = +0.13 \, \text{V} (Note: The sign of the potential is reversed when flipping the reaction.)

To find the standard cell potential (EcellE^\circ_{\text{cell}}), we add the reduction potential of the cathode and the oxidation potential of the anode:

Ecell=Ecathode+EanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode}} Ecell=0.77V+0.13VE^\circ_{\text{cell}} = 0.77 \, \text{V} + 0.13 \, \text{V} Ecell=0.90VE^\circ_{\text{cell}} = 0.90 \, \text{V}

Since the standard cell potential (EcellE^\circ_{\text{cell}}) is positive (+0.90 V), the reaction is spontaneous under standard conditions.

Would you like more details or have any questions?

Here are some related questions:

  1. How do standard reduction potentials help determine the spontaneity of a reaction?
  2. What is the significance of the standard hydrogen electrode (SHE) in electrochemistry?
  3. How can you calculate the Gibbs free energy change (ΔG\Delta G^\circ) for an electrochemical cell?
  4. What is the Nernst equation and how does it apply to non-standard conditions?
  5. How do you determine which half-reaction to reverse when calculating the standard cell potential?
  6. What is the difference between galvanic cells and electrolytic cells?
  7. How does concentration affect the cell potential in a galvanic cell?
  8. What are the practical applications of galvanic cells in everyday life?

Tip: Remember, the standard cell potential is calculated under standard conditions, which are 1 M concentration for aqueous species, 1 atm pressure for gases, and 25°C temperature.

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Math Problem Analysis

Mathematical Concepts

Electrochemistry
Reduction potentials
Standard cell potential

Formulas

Nernst equation

Theorems

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Suitable Grade Level

Advanced High School