Question

Consider the following reaction. Mg(s) + 2VO2+ (aq)+ 4 H+(aq) → Mg2+(aq) + 2 VO2+(aq) + H2O(l)E⁰cell = 3.37 V.a.Calculate the ΔG⁰∘ for the reaction.Express your answer to three significant figures.b. Calculate the K for the reaction.Express your answer to three significant figures.

Answer 1

(a) The value of Gibbs free energy for the reaction is
`-6.50* 10^5J/mol`

(b) The value of
`K` at 298 is,
`8.28* 10^(113)`

Explanation :

The given cell reaction is:

`Mg(g)+2VO_2^(+)(aq)+4H^+(aq)\rightarrow Mg^(2+)(aq)+2VO^(2+)(aq)+2H_2O(l)`

The half reaction will be:

Reaction at anode (oxidation) :
`Mg\rightarrow Mg^(2+)+2e^-`

Reaction at cathode (reduction) :
`2VO_2^(+)+4H^++2e^-\rightarrow 2VO^(2+)+2H_2O`

First we have to calculate the Gibbs free energy.

Formula used :

`\Delta G^o=-nFE^o`

where,

`\Delta G^o` = Gibbs free energy = ?

n = number of electrons = 2 (from the reaction)

F = Faraday constant = 96500 C/mole

`E^o` = standard e.m.f of cell = 3.37 V

Now put all the given values in this formula, we get the Gibbs free energy.

`\Delta G^o=-(2* 96500* 3.37)=-650410J/mol=-6.50* 10^5J/mol`

Thus, the value of Gibbs free energy for the reaction is
`-6.50* 10^5J/mol`

Now we have to calculate the value of equilibrium constant.

Formula used :

`\Delta G^o=-2.303* RT* \log K`

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature =
`25^oC=273+25=298K`

K = equilibrium constant = ?

Now put all the given values in this formula, we get the value of
`K`

`-6.50* 10^5J/mole=-2.303* (8.314 J/K/mole)* (298K)* \log K`

`\log K=113.918`

`K=8.28* 10^(113)`

Therefore, the value of
`K` at 298 is,
`8.28* 10^(113)`