Given the table below of half-reaction reduction potentials, calculate the standard cell potential, E°cell for the given reaction. 3 Sn4+ (aq) + 2 Cr (s) ---> 2 Cr3+ (aq) +3 Sn2+ (aq) Half-reaction E° (V) F2 (g) + 2 e ---> 2 F (aq) +2.87 Br2 (1) + 2 e ---> 2 Br (aq) +1.065 Fe3+ (aq) + e -- -> Fe2+ (ag) +0.771 Sn+ (aq) + 2 e ---> Sn2* (aq) +0.154 Fe2+ (aq) + 2 e ---> Fe (s) -0.440 Cr3+ (ag) + 3 e ---> Cr (s) -0.74 -0.59 +0.89 +2.53 +1.94 -1.02
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![**Calculation of Standard Cell Potential Based on Half-Reaction Reduction Potentials**
This section provides information on how to calculate the standard cell potential (E° cell) for a given redox reaction using the provided half-reaction reduction potentials. Given is the redox reaction:
\[ 3 \text{Sn}^{4+} (\text{aq}) + 2 \text{Cr} (\text{s}) \rightarrow 2 \text{Cr}^{3+} (\text{aq}) + 3 \text{Sn}^{2+} (\text{aq}) \]
**Half-Reaction Reduction Potentials:**
| Half-reaction | E° (V) |
|----------------------------------------|----------|
| F₂(g) + 2 e⁻ → 2 F⁻ (aq) | +2.87 |
| Br₂(l) + 2 e⁻ → 2 Br⁻ (aq) | +1.065 |
| Fe³⁺(aq) + e⁻ → Fe²⁺(aq) | +0.771 |
| Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) | +0.154 |
| Fe²⁺(aq) + 2 e⁻ → Fe(s) | -0.440 |
| Cr³⁺(aq) + 3 e⁻ → Cr(s) | -0.74 |
**Explanation of Reaction Calculation:**
To calculate the standard cell potential for this reaction, follow these steps:
1. Identify the half-reactions involved in the overall reaction.
2. Use the reduction potential values:
- \( \text{Sn}^{4+} \) is reduced to \( \text{Sn}^{2+} \): \( +0.154 \, \text{V} \).
- \( \text{Cr} \) is oxidized to \( \text{Cr}^{3+} \): Since this half-reaction is given as a reduction, flip its sign when used as an oxidation process: \( -(-0.74) = +0.74 \, \text{V} \).
3. Calculate the E° cell by using the formula:
\[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83e4d0e0-41e1-45c9-8283-0ae60fe430d6%2F049c0e25-a8c0-403c-9f2b-5261b53b052d%2Fx2us0fe_processed.jpeg&w=3840&q=75)
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