Study "CHEM (EXAM 4)", a flashcard deck on FORKSAI.
Balance the following redox reaction in acidic solution using the half-reaction method: MnO₄⁻(aq) + C₂O₄²⁻(aq) → Mn²⁺(aq) + CO₂(g).
First, separate into half-reactions: Oxidation: C₂O₄²⁻ → CO₂; Reduction: MnO₄⁻ → Mn²⁺. Balance atoms other than O and H, then O with H₂O, then H with H⁺, and finally charge with electrons. The balanced equation is 2MnO₄⁻(aq) + 5C₂O₄²⁻(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l).
Define the terms 'electrochemical cell', 'voltaic (galvanic) cell', and 'electrolytic cell', and distinguish between them based on spontaneity and energy conversion.
An electrochemical cell is a device that either generates electrical energy from a chemical reaction or uses electrical energy to drive a non-spontaneous chemical reaction. A voltaic (galvanic) cell is a type of electrochemical cell that generates electricity from a spontaneous chemical reaction. An electrolytic cell uses electrical energy to drive a non-spontaneous chemical reaction.
A galvanic cell is constructed with a zinc electrode in a 1.0 M Zn(NO₃)₂ solution and a copper electrode in a 1.0 M Cu(NO₃)₂ solution. Sketch this cell, identifying the anode, cathode, salt bridge, and the direction of electron flow.
The sketch would show two beakers connected by a salt bridge. The zinc electrode (anode) is in the Zn(NO₃)₂ solution, and the copper electrode (cathode) is in the Cu(NO₃)₂ solution. Electrons flow from the zinc electrode (anode) through an external wire to the copper electrode (cathode). The salt bridge allows ion migration to maintain charge neutrality.
A current of 0.50 A flows through a wire for 30 minutes. Calculate the total charge that passes through the wire in Coulombs.
Charge (Q) is calculated as current (I) multiplied by time (t). First, convert time to seconds: 30 minutes * 60 seconds/minute = 1800 seconds. Then, Q = I * t = 0.50 A * 1800 s = 900 C. Therefore, 900 Coulombs of charge pass through the wire.
Explain the significance of the standard hydrogen electrode (SHE) in determining standard electrode potentials for other half-reactions.
The standard hydrogen electrode (SHE) serves as the reference electrode with an assigned standard electrochemical potential of 0.00 V. By connecting any other half-cell to the SHE, the measured cell potential directly corresponds to the standard electrode potential of that specific half-reaction, allowing for a standardized comparison of reduction potentials.
Given the standard reduction potentials: Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V, and Fe³⁺(aq) + e⁻ → Fe²⁺(aq), E° = +0.77 V. Calculate the standard cell potential for a galvanic cell involving these two half-reactions and predict the direction of spontaneity.
To form a galvanic cell, one half-reaction must be reversed to act as oxidation. Since Ag⁺ has a higher reduction potential, it will be reduced, and Fe²⁺ will be oxidized. The oxidation half-reaction is Fe²⁺(aq) → Fe³⁺(aq) + e⁻, with E° = -0.77 V. The standard cell potential E°cell = E°cathode - E°anode = +0.80 V - (+0.77 V) = +0.03 V. Since E°cell is positive, the reaction is spontaneous in this direction.
For a redox reaction, if the standard change in free energy (ΔG°rxn) is -150 kJ/mol, calculate the standard potential for the electrochemical cell (E°) at 25°C. Assume n = 2 for the reaction. (Faraday's constant F = 96485 C/mol).
The relationship is ΔG°rxn = -nFE°. Rearranging for E°: E° = -ΔG°rxn / (nF). E° = -(-150,000 J/mol) / (2 mol e⁻ * 96485 C/mol e⁻) = 150,000 J / 192,970 C = 0.777 V. The standard cell potential is approximately +0.78 V.
A concentration cell is set up with two Ag/Ag⁺ half-cells. One half-cell has [Ag⁺] = 0.10 M and the other has [Ag⁺] = 0.0010 M. Calculate the cell potential (E) at 25°C using the Nernst equation. (R = 8.314 J/(mol·K), F = 96485 C/mol).
For a concentration cell, E°cell = 0 V. The Nernst equation is E = E° - (RT/nF)lnQ. Here, n=1. Q = [Ag⁺]anode / [Ag⁺]cathode. The more dilute solution is the anode. Q = 0.0010 M / 0.10 M = 0.01. E = 0 - (8.314 * 298 / (1 * 96485))ln(0.01) = -0.0257 * (-4.605) = 0.118 V. The cell potential is approximately 0.12 V.
Define a 'nuclide' and distinguish it from an 'isotope'. Provide an example for each.
A nuclide is a general term for an atom characterized by a specific number of protons and neutrons in its nucleus. For example, carbon-12 (¹²C) is a nuclide. Isotopes are nuclides of the same element (same number of protons) but with different numbers of neutrons. Carbon-12 and carbon-14 (¹⁴C) are isotopes of carbon.
Balance the following redox reaction in basic solution: Cr(OH)₃(s) + ClO⁻(aq) → CrO₄²⁻(aq) + Cl⁻(aq).
First, balance in acidic solution then convert to basic. The balanced acidic equation is 2Cr(OH)₃(s) + 3ClO⁻(aq) + 4H⁺(aq) → 2CrO₄²⁻(aq) + 3Cl⁻(aq) + 5H₂O(l). To convert to basic, add 4OH⁻ to both sides to neutralize H⁺. The final balanced basic equation is 2Cr(OH)₃(s) + 3ClO⁻(aq) + 4OH⁻(aq) → 2CrO₄²⁻(aq) + 3Cl⁻(aq) + 5H₂O(l).