Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

20. Electrochemistry

Standard Reduction Potentials

General Chemistry

20. Electrochemistry

Standard Reduction Potentials

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Multiple Choice

What is the value (in V) of E°cell for the following reaction: Sn²⁺ (aq) + Fe (s) → Sn (s) + Fe²⁺ (aq)?

0.14 V

0.34 V

-0.34 V

-0.14 V

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Verified step by step guidance

Identify the half-reactions involved in the overall reaction. The oxidation half-reaction is Fe (s) → Fe²⁺ (aq) + 2e⁻, and the reduction half-reaction is Sn²⁺ (aq) + 2e⁻ → Sn (s).

Look up the standard reduction potentials (E°) for each half-reaction. For Fe²⁺/Fe, E° = -0.44 V, and for Sn²⁺/Sn, E° = -0.14 V.

Determine which half-reaction is the reduction and which is the oxidation. The half-reaction with the more positive E° value will be the reduction. In this case, Sn²⁺/Sn is the reduction, and Fe/Fe²⁺ is the oxidation.

Calculate the standard cell potential (E°cell) using the formula: E°cell = E°(cathode) - E°(anode). Here, the cathode is Sn²⁺/Sn and the anode is Fe/Fe²⁺.

Substitute the standard reduction potentials into the formula: E°cell = (-0.14 V) - (-0.44 V). Simplify the expression to find the E°cell value.