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: $3\times 10^{35}$

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

$\Delta G=-2\times 96500\times (-1.05)=202650J$

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

$\Delta G=-2.303RTlog K$

$+202650=-2.303\times 8.314\times 298\times logK$

$K=3\times 10^{35}$

Thus the value of the equilibrium constant at $25^0C$ is $3\times 10^{35}$