Question

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Ag(s) and Ni2+(aq) to give Ni(s) and Ag+(aq) Use the reduction potential values for Ag+(aq) of +0.80 V and for Ni2+(aq) of -0.25 V Give your answer using E-notation with NO decimal places

Answer 1

Answer:
`3* 10^(35)`

Step-by-step explanation:

The balanced chemical equation will be:

`2Ag(s)+Ni^(2+)(aq)\rightarrow 2Ag^(+)(aq)+Ni(s)`

Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.

`E^0=E^0_(cathode)- E^0_(anode)`

Where both
`E^0` are standard reduction potentials.

`E^0_([Ag^(+)/Mg])=+0.80V`

`E^0_([Ni^(2+)/Ni])=-0.25V`

`E^0=E^0_([Ni^(2+)/Ni])- E^0_([Ag^(+)/Ag])`

`E^0=-0.25-(+0.80V)=-1.05V`

The standard emf of a cell is related to Gibbs free energy by following relation:

`\Delta G=-nFE^0`

`\Delta G` = gibbs free energy

n= no of electrons gained or lost =?

F= faraday's constant

`E^0` = standard emf

`\Delta G=-2* 96500* (-1.05)=202650J`

The Gibbs free energy is related to equilibrium constant by following relation:

`\Delta G=-2.303RTlog K`

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin =
`25^0C=25+273=298K`

K = equilibrium constant

`\Delta G=-2.303RTlog K`

`+202650=-2.303* 8.314* 298* logK`

`K=3* 10^(35)`

Thus the value of the equilibrium constant at
`25^0C` is
`3* 10^(35)`