Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°c. sn(s) | sn2+(aq, 0.022 m) || ag+(aq, 2.7 m) | ag(s)

a. -0.83 v
b. +1.01 v
c. -0.66 v
d. +1.31 v
e. +0.01 v

b. + 1.01 V

Explanation:

Sn(s) ⇒ Sn2+ + 2e⁻     Eox = 0.14 V

Ag₊ + e⁻⇒                   Ered = 0.80 V

Sn(s) + Ag⁺(aq) ⇒ Sn2⁺(aq)  + Ag(s)    Eºcell = 0.14 V + 0.80 V =  0.94 V

This problem is solved by using the Nernst equation:

Ecell = Eºcell -  (0.0592 / n) log Q

since we are not at standard condition given that the concentrations are not at  1 M. So we will calculate  Q first

Q =  ( Sn²⁺ ) / ( Ag⁺) = 0.022/2.7 = 0.0081

Ecell = 0.94 V -(0.0592/2) x log (0.0081) = 1.00 V

може ли да зареждате повече информация, моля