Chemistry

      Q1. In a galvanic cell based on the half reactions Ni²⁺(aq) + 2e⁻ → Ni(s) E° = -0.250 V 2H⁺(aq) + 2e⁻ → H₂ E° = 0.000 V the nickel compartment contains a nickel electrode in a solution where [Ni²⁺] = 1.00 × 10⁻² M, and the hydrogen compartment contains a platinum electrode, P_H₂ = 1.00 atm, and a weak acid, HA, at an initial concentration of 1.00 M. If the observed cell potential is 0.170 V at 25.0 °C, calculate the Ka value for the weak acid. Q2. A voltaic cell using Cu²⁺/Cu and Al³⁺/Al half-cells is set up at standard conditions, and each compartment has a volume of 225 mL. What is the [Al³⁺] after the cell has delivered 0.120 A for 59.0 hours at 25.0 °C? (E° for Cu²⁺/Cu = 0.340 V and E° for Al³⁺/Al = -1.660 V.) Q3. Complete the stepwise process of balancing the following reaction in acidic conditions. Cr₂O₇²⁻(aq) + HNO₂(aq) → Cr³⁺(aq) + NO₃⁻(aq) a. The first two steps of balancing a redox reaction are to separate the reaction into half reactions and then balance all atoms EXCEPT H and O. Balance this half reaction for all atoms except H and O. Cr₂O₇²⁻(aq) → Cr³⁺(aq) b. The first two steps of balancing a redox reaction are to separate the reaction into half reactions and then balance all atoms EXCEPT H and O. Balance this half reaction for all atoms except H and O. HNO₂(aq) → NO₃⁻(aq) c. The third step is to balance the oxygen atoms using water . Balance this half reaction with H₂O. Cr₂O₇²⁻(aq) → 2 Cr³⁺(aq) + H₂O(l) d. The third step is to balance the oxygen atoms using water . Balance this half reaction with H₂O. HNO₂(aq) + H₂O(l) → NO₃⁻(aq) e. The fourth step is to balance the hydrogen atoms using H⁺. Balance this half reaction with H⁺. Cr₂O₇²⁻(aq) + H⁺→ 2 Cr³⁺(aq) + 7 H₂O(l) f. The fourth step is to balance the hydrogen atoms using H⁺. Balance this half reaction with H⁺. HNO₂(aq) + H₂O(l) → NO₃⁻(aq) + H⁺ g. The fifth step is to balance the charge using electrons. Balance this half reaction with the addition of e⁻. Cr₂O₇²⁻(aq) + 14 H⁺(aq) + e⁻ → 2 Cr³⁺(aq) + 7 H₂O(l) h. The fifth step is to balance the charge using electrons. Balance this half reaction with the addition of e⁻. HNO₂(aq) + H₂O(l) → NO₃⁻(aq) + 3 H⁺(aq) + e⁻ i. The sixth step of balancing redox reactions is to multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other. For the two half reactions shown below, how should they be scaled to balance the electrons? Cr₂O₇²⁻(aq) + 14H⁺ + 6e⁻→ 2Cr³⁺(aq) + 7H₂O(l) HNO₂(aq) + H₂O(l) → NO₃⁻(aq) + 3H⁺ + 2e⁻ j. The last step is to combine the two half reactions (shown below) into one complete reaction. Cr₂O₇²⁻(aq) + 14H⁺ + 6e⁻→ 2Cr³⁺(aq) + 7H₂O(l) 3HNO₂(aq) + 3H₂O(l) → 3NO₃⁻(aq) + 9H⁺ + 6e⁻  

E = E° - (RT/nF) ln(Q)

Q = [Ni²⁺] / [H²]²

Q = (1.00 × 10⁻² M) / (1.00 atm)^2

Q = 1.00 × 10⁻² M

0.170 V = -0.250 V - (8.314 J/mol·K / (2 * 96,485 C/mol)) * (298.15 K) * ln(1.00 × 10⁻² M)

ln(1.00 × 10⁻² M) = -0.792

1.00 × 10⁻² M = e⁻⁰⁷⁹²

Ka = [H⁺]^2 / [HA]

Ka = (e⁻⁰⁷⁹²)^2 / 1.00 M

Ka = 6.31 × 10⁻¹⁶

n = tI / F

59.0 hours * 3600 seconds/hour = 212,400 seconds

n = (212,400 seconds) * (0.120 A) / (96,485 C/mol)

n = 2.60 moles

Al(s) → Al³⁺(aq) + 3e⁻
Cu²⁺(aq) + 2e⁻ → Cu(s)

[Al³⁺] = 0.867 moles / 0.225 L

[Al³⁺] = 3.86 M

[Al³⁺] = 3.86 M - 2.60 M

[Al³⁺] = 1.26 M

Cr₂O₇²⁻(aq) + 6HNO