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)

To determine the standard cell potential (\(E^\circ_{\text{cell}}\)) 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. \( \text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq) \quad E^\circ = +0.77 \, \text{V} \)
2. \( \text{Pb}^{2+} (aq) + 2 e^- \rightarrow \text{Pb} (s) \quad E^\circ = -0.13 \, \text{V} \)

The overall reaction is:

\[ \text{Pb}^{2+} (aq) + 2 \text{Fe}^{2+} (aq) \rightarrow 2 \text{Fe}^{3+} (aq) + \text{Pb} (s) \]

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

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

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):
\[ \text{Fe}^{3+} (aq) + e^- \rightarrow \text{Fe}^{2+} (aq) \quad E^\circ_{\text{red}} = +0.77 \, \text{V} \]

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

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

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 0.77 \, \text{V} + 0.13 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.90 \, \text{V} \]

Since the standard cell potential (\(E^\circ_{\text{cell}}\)) 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 (\(\Delta G^\circ\)) 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.

Learn how to calculate the standard cell potential for a galvanic cell reaction using reduction potentials. Determine if the reaction is spontaneous under standard conditions. Detailed explanation with examples and formulas.

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