Question

What are the steps you would use to work this out?

What is the voltage for a cell constructed of Ni(s)in 1.0 x 10-7M Ni+2and Ag(s)in 1.0 x 10-5M Ag+?

Answer 1

THe voltage is the measure of the cell potential of an electrochcemeical cell

Now the given electrochemical cell is Ni(s)/Ni⁺²// Ag(s)/Ag⁺   

Every electrochemical cell is one redox reaction. That is both reduction and oxidation takes place in one cell. So according to type of reaction, we can devide this electrochemical cell into two half cells

1- Oxidation half - loss of electron take place   Ni(s)/Ni⁺² = Ni(s) --------------> Ni⁺² + 2e

2- Reduction half - gain of electron take place   Ag(s)/Ag⁺ = Ag⁺ + e --------------> Ag(s)

Now the Standard reduction potential value (E⁰red) of the two cells from any reference table will show in which cell the actual reaction takes place. The cell with higher E⁰red value is the actual reduction half cell

Now For Ni(s)/Ni⁺² = Ni⁺²​​​​​​​ + 2e --------------> Ni(s) E⁰red = -0.23 V

Again for   Ag⁺ + e --------------> Ag(s)   E⁰red = 0.80 V

Thus form these two E⁰red values, we can clearly see that actual reduction takes place at Ag and oxidation will take place at - Ni

Thus Oxidation half Ni(s) --------------> Ni⁺²​​​​​​​ + 2e E⁰red = -0.23 V

  Reduction half 2Ag⁺ + 2e --------------> 2Ag(s) E⁰red = 2 *0.80 = 1.80 V

Thus Overall reaction   Ni(s) + 2Ag⁺ -------------> Ni⁺²​​​​​​​ + 2Ag(s)   E⁰cell = Higher - lower

= 1.80 V - (-0.23 V)

= 2.03‬ V

Now the total cell potential (Ecell) is calculated by using Nernst Equation

Ecell = E⁰cell - (0.0592/n) * log[Ox]/[Red]

where n = number of electrons involved

  [Ox] = concentration of oxydising species

[Red] = concentration of Reducing specied

Thus

Ecell = E⁰cell - (0.0592/n) * log[Ni⁺²]/[Ag⁺]

Ecell = 2.03‬ V - (0.0592 / 2) * log[1.0 x 10^-7]/[1.0 x 10^-5]

=  2.03‬ V - 0.0296 * log[ 10^-2]

= 2.03‬ V - 0.0296 *(-2)

= 2.03‬ V + 0.0592

= 2.0892 V

Thus the total voltage obtained form the cell Ni(s)in 1.0 x 10-7M Ni+2and Ag(s)in 1.0 x 10-5M Ag+ is 2.0892 V