Question

What is the voltage of the following cell? Cr ⁄ Cr⁺³(aq) (0.010 M) ⁄⁄ Ag⁺(aq) (0.00010 M) ⁄ Ag

The unbalanced reaction is: Cr(s) + Ag⁺(aq) → Cr⁺³(aq) + Ag(s) ; E^o_cell = +1.53 V

Answer 1

The balanced reaction is :

Cr(s) + 3Ag⁺(aq) → Cr⁺³(aq) + 3Ag(s) ; E^o_cell = +1.53 V

0.00010M 0.010 M

According to Nernst Equation ,

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

   = E^o_cell - (0.059 / n) log ([Cr³⁺] / [Ag⁺]³ )

Where

E = electrode potential of the cell = ?

E^o_cell = standard electrode potential = +1.53 V

n = number of electrons involved in the reaction = 3

[Cr³⁺] = 0.010 M

[Ag⁺] = 0.00010 M

Plug the values we get

E = E^o_cell - (0.059 / n) xlog ([Cr³⁺] / [Ag⁺]³ )

   = +1.53 - (0.059 / 3 ) x log ( 0.010 / 0.00010 )

= 1.491 V