Question

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate ΔG for this reaction.

*** Please explain the reactions since I’m very confused as to wich side I should put the electrons.
Ex: Cu-> Cu2+ + 2e

Answer 1

Answer : The
`\Delta G` for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

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

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation :
`Cu\rightarrow Cu^(2+)+2e^-`

Reduction :
`2Ag^++2e^-\rightarrow 2Ag`

Now 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 to balance the reaction = 2

F = Faraday constant = 96500 C/mole

`E^o` = standard e.m.f of cell = 0.46 V

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

`\Delta G^o=-(2* 96500* 0.46)=-88780J/mole`

Therefore, the
`\Delta G` for this reaction is, -88780 J/mole.