Question

Calculate the cell potential, at 25^oC, for the reaction

3 Zn(s) + 2 Cr⁺³(aq)[0.010 M] --> 3 Zn⁺²(aq)[0.020 M] + 2 Cr(s)

given,

Cr⁺³(aq) + 3e^- --> Cr(s) . . . . . . . E^o= -0.74 V

Zn⁺²(aq) + 2e^- --> Zn(s) . . . . . . E^o = -0.76 V

Question options:

	-0.03 V
	+0.01 V
	+ 0.03 V
	-0.01 V
	+ 0.02 V

Answer 1

3 Zn(s) + 2 Cr⁺³(aq)[0.010 M] 3 Zn⁺²(aq)[0.020 M] + 2 Cr(s)

So standard potential of the cell , E^o = E^o_cathode - E^o_anode

                                                      = E^o_Zn2+/Zn - E^o_Cr3+/Cr

                                                      = -0.74 - (-0.76) V

                                                      = +0.02 V

According to Nernst Equation ,

E = E^o - (0.059 / n) log ([Products] / [reactants] )

   = E^o - (0.059 / n) log ([Zn²⁺] / [Cr³⁺] )

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +0.02 V

n = number of electrons involved in the reaction = 6

[Cr³⁺] = 0.010 M

[Zn²⁺] = 0.020 M

Plug the values we get

E = E^o - (0.059 / n) xlog ([Zn²⁺] / [Cr³⁺] )

   = +0.02 - (0.059 / 6 ) x log ( 0.020 / 0.010 )

   = +0.017 V

   ~ +0.02