Respuesta :
Answer: The [tex]\Delta G^o[/tex] for the given reaction is [tex]-5.23\times 10^5J[/tex]
Explanation:
For the given chemical reaction:
[tex]2Na(s)+2H_2O(l)\rightarrow H_2(g)+2OH^-(aq.)+2Na^+(aq.)[/tex]
Half reactions for the given cell follows:
Oxidation half reaction: [tex]Na(s)\rightarrow Na^{+}+e^-;E^o_{Na^{+}/Na}=-2.71V[/tex] ( × 2)
Reduction half reaction: [tex]2H^{+}+2e^-\rightarrow H_2(g);E^o_{H^{+}/H}=0.00V[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.00-(-2.71)=2.71V[/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 = 2
F = Faradays constant = 96500 C
[tex]E^o_{cell}[/tex] = standard cell potential = 2.71 V
Putting values in above equation, we get:
[tex]\Delta G^o=-2\times 96500\times 2.71=-523030J=-5.30\times 10^5J[/tex]
Hence, the [tex]\Delta G^o[/tex] for the given reaction is [tex]-5.23\times 10^5J[/tex]
