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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

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

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.101QE: At 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction...
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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^\
Transcribed Image Text:**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^\
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