Question

Questions are as shown. NO additional information was given.

1. Calculate the f.e.m. for the following reaction at 25^oC Li⁺_(aq) (1.8e-4M) +   2Cl^-_(aq) (6.9e-2M)      →   Li_(s) ( - M) +   Cl_2(g) ( - M)

E_cell = ? (2 decimals accuracy!)

2. Calculate the f.e.m. for the following reaction at 25^oC Au³⁺_(aq) (3.0e-2M) +   2I^-_(aq) (2.2e-3M)      →   Au_(s) ( - M) +   I_2(s) ( - M)

E_cell =? (2 decimals accuracy!)

Answer 1

The standard electrode potential for Li system is Li⁺ + e^- -------> Li ; E^o red = -3.045 V ---(1)

The standard electrode potential for Cl system is Cl₂ + 2e^- --------> 2Cl^- ; E^o red = +1.360 V ---(2)

Since the reduction potential of Cl system is more it undergoes oxidation at anode & Li undergoes reduction at cathode.

The cell reaction is 2Li⁺ + 2Cl^- --------> 2Li + Cl₂

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

                                                      = E^o_Li+/Li - E^o_Cl2/Cl-

= -3.045 -(+1.360)

= -4.405 V

According to Nernst Equation ,

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

   = E^o - (0.059 / n) log (1 / ([Li⁺]²[Cl-]² ))

Where

E = electrode potential of the cell = ?

E^o = standard electrode potential = +1.103 V

n = number of electrons involved in the reaction = 2

[Cl^-] = 6.9x10^-2 M

[Li⁺] = 1.8x10^-4 M

Plug the values we get

E = E^o - (0.059 / n) xlog (1 / ([Li⁺]²[Cl-]² ))

   = -4.405 - (0.059 / 2 ) x log (1 /(1.8x10^-4)² x (6.9x10^-2)²))

   = -4.69 V

Simillarly do the second one