NEETChemistryElectrochemistry
The standard Gibbs energy change ( _ r G^ ) for the following reaction is -868.5 kJ mol ⁻¹ : 2 M(s) + 6 H ⁺( aq ) 2 M ³⁺( aq ) + 3 H ₂( g ) Calculate the standard reduction potential of the M ³⁺/ M half-cell. (Given : 1 F = 96500 C mol ⁻¹ )
Options
- A-1.50 V
- B+1.50 V
- C-3.00 V
- D-4.50 V
Correct answer
A. -1.50 V
Step-by-step solution
For the given balanced reaction, the number of moles of electrons transferred is n = 6 . Using the relation: _ r G^ = -nFE_ cell ^ -868500 J mol ⁻¹ = -6 96500 C mol ⁻¹ E_ cell ^ E_ cell ^ = 868500 579000 V = +1.50 V The cell consists of the M anode and the Standard Hydrogen Electrode (SHE) as the cathode. E_ cell ^ = E_ cathode ^ - E_ anode ^ 1.50 V = E_ H ⁺/ H ₂ ^ - E_ M ³⁺/ M ^ 1.50 V = 0 V - E_ M ³⁺/ M ^ E_ M ³⁺/ M ^ = -1.50 V Answer: -1.50 V