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?

0.35 V

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

(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:  

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