Question

A Mg|Mg2+ || Ni2+|Ni galvanic cell is constructed in which the standard cell voltage is 2.12 V. Calculate the free energy change at 25°C when 0.823 g of Ni plates out, if all concentrations remain at their standard value of 1 M throughout the process. What is the maximum amount of work that could be done by the cell on its surroundings during this experiment?

ΔG° =____________ J

Maximum work =___________________ J

Answer 1

Answer : The value of
`\Delta G^o=-409160J` and maximum work = 409160 J

Explanation :

The given galvanic cell is:

`Mg/Mg^(2+)||Ni^(2+)/Ni`

From the cell we conclude that, magnesium shows oxidation and act a anode and nickel shows reduction and act as cathode.

The balanced two-half reactions will be,

Oxidation half reaction :
`Mg(s)\rightarrow Mg^(2+)(aq)+2e^-`

Reduction half reaction :
`Ni^(2+)(aq)+2e^-\rightarrow Ni(s)`

Thus the overall reaction will be,

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

Now we have to calculate the Gibbs free energy.

Formula used :

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

where,

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

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

`E^o` = standard e.m.f of cell = 2.12 V

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

`\Delta G^o=-(2* 96500* 2.12)=-409160J/mole`

Therefore, the value of
`\Delta G^o=-409160J` and maximum work = 409160 J