Question

What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of the ions in solution were [Cr3+] = 0.75 M and [Fe2+] = 0.25 M at 298K?

Answer 1

0.35 V

Step-by-step explanation:

(a) Standard reduction potentials

E°/V

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

E°/V

2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74

3Fe ⇌ 3Fe²⁺ + 6e-; -0.41

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-

2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)

The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

`E = E^(\circ) - (RT)/(zF)\ln Q`

(a) Data

E° = 0.33 V

R = 8.314 J·K⁻¹mol⁻¹

T = 298 K

z = 6

F = 96 485 C/mol

(b) Calculations:

`Q = \frac{\text{[Fe}^(2+)]^(3)}{ \text{[Cr}^(3+)]^(2)} = (0.25^(3))/( 0.75^(2)) =(0.0156)/(0.562) = 0.0278\\\\E = 0.33 - \left ((8.314 * 298)/(6 * 96485)\right ) \ln(0.0278)\\\\=0.33 -0.00428 * (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}`