Question

PLEASE HELP! A voltaic cell is constructed that uses the following half-cell reactions: Cu+(aq)+e−→Cu(s)I2(s)+2e−→2I−(aq)E∘=0.52VE∘=0.54V. The cell is operated at 298 K with [Cu+]=0.31 M and [I−]= 3.0 M.

a) Determine E for the cell at these concentrations.
d) If [Cu+] was equal to 0.15 M, at what concentration of I− would the cell have zero potential?

Answer 1

Answer:

a) Ecell = 0.074 V

d) [I-] = 0.183 M

Step-by-step explanation:

a) The standard cell potential (E°) can be calculated as:

E°cell = E°reduction (cathode) - E°reduction (anode)

E°cell = 0.54 V - 0.52 V

E°cell = 0.02 V

The Nernst equation can be used to calculate the cell potential (Ecell) at non-standard conditions:

Ecell = E°cell - (RT/nF) × ln(Q)

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), n is the number of electrons transferred (2 in this case), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

The reaction quotient can be calculated as:

Q = [Cu+]/[I-]^2

Substituting the values:

Q = (0.31 M) / (3.0 M)^2 = 0.034 M^-2

Now, we can calculate Ecell:

Ecell = 0.02 V - [(8.314 J/mol·K) / (2 × 96,485 C/mol)] × ln(0.034)

Ecell = 0.02 V - (0.00273 V) × ln(0.034)

Ecell = 0.02 V - (-0.054 V)

Ecell = 0.074 V

Therefore, the cell potential at these concentrations is 0.074 V.

d) To find the concentration of I- at which the cell potential is zero, we can use the equation:

Ecell = E°cell - (RT/nF) × ln(Q)

Setting Ecell to zero and solving for [I-]:

0 = 0.02 V - (0.0592 V / 2) × log([Cu+]/[I-]^2)

0.02 V = 0.0296 V × log([Cu+]/[I-]^2)

log([Cu+]/[I-]^2) = 0.67

[Cu+]/[I-]^2 = 4.48

0.15 M / [I-]^2 = 4.48

[I-]^2 = 0.0336

[I-] = sqrt(0.0336) = 0.183 M

Therefore, the concentration of I- at which the cell potential is zero, with [Cu+] equal to 0.15 M, is 0.183 M.