Chemistry

Full Answer Section

We know the following values:

• E = 0.170 V
• E° = -0.250 V
• R = 8.314 J/mol·K
• T = 298.15 K
• n = 2
• F = 96,485 C/mol

We can calculate Q by using the concentrations of the nickel and hydrogen ions:

Q = [Ni²⁺] / [H²]²

Q = (1.00 × 10⁻² M) / (1.00 atm)^2

Q = 1.00 × 10⁻² M

Plugging all of the known values into the Nernst equation, we get:

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

Therefore, the Ka value for the weak acid is 6.31 × 10⁻¹⁶.

Q2

To calculate the [Al³⁺] after the cell has delivered 0.120 A for 59.0 hours at 25.0 °C, we can use the following equation:

n = tI / F

where:

• n is the number of moles of electrons transferred
• t is the time in seconds
• I is the current in amperes
• F is the Faraday constant

We need to convert the time from hours to seconds:

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

Plugging all of the known values into the equation, we get:

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

n = 2.60 moles

The following equation shows the balanced half-reactions for the voltaic cell:

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

From the balanced half-reactions, we can see that 3 moles of electrons are transferred for every mole of Al(s) that is oxidized. Therefore, 2.60 moles of electrons correspond to the oxidation of 0.867 moles of Al(s).

The initial volume of the Al³⁺ compartment is 225 mL. Therefore, the initial concentration of Al³⁺ is:

[Al³⁺] = 0.867 moles / 0.225 L

[Al³⁺] = 3.86 M

After 0.120 A has been delivered for 59.0 hours, the concentration of Al³⁺ will be:

[Al³⁺] = 3.86 M - 2.60 M

[Al³⁺] = 1.26 M

Therefore, the [Al³⁺] after the cell has delivered 0.120 A for 59.0 hours at 25.0 °C is 1.26 M.

Q3

The balanced reaction in acidic conditions is:

Cr₂O₇²⁻(aq) + 6HNO

Sample Solution

Q1

To calculate the Ka value for the weak acid, we can use the Nernst equation:

E = E° - (RT/nF) ln(Q)

where:

• E is the cell potential
• E° is the standard cell potential
• R is the gas constant
• T is the temperature in Kelvin
• n is the number of electrons transferred in the reaction
• F is the Faraday constant
• Q is the reaction quotient