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

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.