Question

44. Determine the Cell potential for Cr(3+)(aq) + C12(aq) - -> Cr207(2-)(aq) CI-(aq) Assume 1.0 M and all partial pressures are 1.0 atm. a. 0.06 V b. 1.50 V с. 0.03 V 45. Determine the Cell potential for Cu(2+)(aq) + Mg(s) --> Mg(2+)(ag) + Cu(s) a. 8.83 V C. 4.45 V

Answer 1

Solution:

The emf of a cell ( E°cell) is calculated from standard reduction potentials of two half cells as:

E°cell = E°reduction half cell - E° oxidation half cell

44) For the given reaction:

Cr3+ + Cl2 = Cr2O7^2- + Cl-

Oxidation half cell reaction is:

2Cr3+ = Cr2O7^2- ( Cr6+)

The standard reduction potential is:

E° ( Cr6+ / Cr3+) = +1.33 V

Reduction half cell reaction is:

Cl2 + 2 e- = 2Cl-

E° (Cl2/Cl-) = +1.36 V

Hence, E° cell = E°( Cl2/ Cl-) - E° ( Cr6+ / Cr3+)

= 1.36 V - 1.33 V = = 0.03 V

Therefore, correct option is C

45) For the given reaction:

Cu2+ + Mg = Mg2+ + Cu

The oxidation half cell :

Mg -2e-  = Mg2+

E° ( Mg2+/Mg) = -2.37 V

The reduction half cell:

Cu2+ + 2e- = Cu

E°(Cu2+/Cu) = +0.34 V

Hence, E°cell = E° ( Cu2+/Cu) - E°(Mg2+/ Mg)

= +0.34 - (- 2.37 ) = 0.34 + 2.37 = +2.71 V

Therefore, option is B.