Using the relationship given in part 1, calculate Ecell if [Cu2+] = 1.753 M and [Cr3+] = 0.00143 M, assuming that the temperature remains at 25°C. Give your answer to two places after the decimal.

Question

Using the relationship given in part 1, calculate E_cell if [Cu²⁺] = 1.753 M and [Cr³⁺] = 0.00143 M, assuming that the temperature remains at 25°C. Give your answer to two places after the decimal.

(R = 8.314 J/mol·K and F = 9.65 × 10⁴ C/mol e^–)

An electrochemical cell is set up at 25°C based on the overall reaction represented by the balanced equation shown below. The cell initially has [Cr³⁺] = [Cu²⁺] and E°_cell = +1.08 V.

6 moles of electrons are transferred per reaction cycle

3Cu²⁺(aq) + 2Cr (s) --> 3 Cu (s) + 2Cr³⁺(aq)

Answer 1

3Cu²⁺(aq) + 2Cr (s) --> 3 Cu (s) + 2Cr³⁺(aq) ... Eºcell = +1.08 V

Use the Nernst equation to find Ecell under the conditions described:

Ecell = Eºcell - RT/nF ln Q and at 25ºC and converting to log₁₀ we have ...

Ecell = Eºcell - 0.0591/n log Q

Ecell = ?

Eºcell = 1.08

n = 6 moles e-

F = 96,500 C mole e-

Q = [Cr³⁺]² / [Cu²⁺]³ = (0.00143)² / (1.753)³ = 3.796x10^-7

Ecell = 1.08 - 0.0591/6 log 3.796x10^-7

Ecell = 1.08 - 0.00985 x -6.42

Ecell = 1.08 + 0.063

Ecell = 1.14 V

(be sure to check all of the math for possible errors)