Question

A concentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two Zn2+ solutions, 0.129 M and 0.427 M, and all other materials needed for a galvanic cell. What will the potential of this cell be when the cathode concentration of Zn2+ has changed by 0.047 M at 291 K?

Zn2+ + 2 e- ? Zn Eo = -0.761 V

Answer 1

Step-by-step explanation:

The given reaction at cathode will be as follows.

At cathode:
`Zn^(2+) + 2e^(-) \rightarrow Zn`,
`E_(o)` = -0.761 V

At anode:
`Zn \rightarrow Zn^(2+) + 2e^(-)`,
`E_(o)` = 0.761

Therefore, net reaction equation will be as follows.

`Zn^(2+) + Zn \rightarrow Zn + Zn^(2+)`

Initial: 0.129 - - 0.427

Change: -0.047 - - -0.047

Equilibrium: (0.129 - 0.047) (0.427 - 0.047)

= 0.082 = 0.38

As
`E^(o)_(cell)` for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

E_{cell} =
`E^(o)_(cell) - (RT)/(nF) ln ([Zn^(2+)]_(products))/([Zn^(2+)]_(reactants))`

=
`0 - (8.314 atm L/mol K * 291 K)/(2 * 96500) ln (0.38)/(0.082)`

= 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.