Answer: 0.124 V
Explanation:
[tex]X(s)+Y^{4+}(aq)\rightarrow X^{4+}(aq)+Y(s)[/tex]
To calculate standard Gibbs free energy, we use the equation:
[tex]\Delta G^o=-nFE^o_{cell}[/tex]
Where,
n = number of electrons transferred = 4
F = Faradays constant = 96500 C
[tex]E^o_{cell}[/tex] = standard cell potential = ?
Putting values in above equation, we get:
[tex]-48.0\times 1000=-4\times 96500\times E^0{cell}[/tex]
[tex]E^o_{cell}=0.124[/tex]
Thus the standard potential E°, for this reaction is 0.124 V
The standard cell potential is 0.12 V.
We know that;
ΔG° = -nFE° cell
Where;
ΔG° = Standard free energy change
n = Number of moles of electrons = 4
F = Faraday's constant = 96500 C
E° cell = Standard cell potential
Making E° cell the subject of the formula;
E° cell = ΔG°/-nF
E° cell = − 48.0 × 10^3J/-(4 × 96500)
E° cell = 0.12 V
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