Respuesta :
Answer: The cell voltage of the given chemical reaction is 2.74 V
Explanation:
For the given chemical reaction:
[tex]Sn^{2+}(aq.)+Ba(s)\rightarrow Ba^{2+}(aq.)+Sn(s)[/tex]
Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.
Oxidation half reaction: [tex]Ba(s)\rightarrow Ba^{2+}(aq.)+2e^-;E^o_{Ba^{2+}/Ba}=-2.90V[/tex]
Reduction half reaction: [tex]Sn^{2+}(aq.)+2e^-\rightarrow Sn(s);E^o_{Sn^{2+}/Sn}=-0.14V[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as 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]
[tex]E^o_{cell}=-0.14-(-2.90)=2.76V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]}{[Ba^{2+}]}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = ?V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +2.76 V
n = number of electrons exchanged = 2
[tex][Ba^{2+}]=5.15M[/tex]
[tex][Sn^{2+}]=1.59M[/tex]
Putting values in above equation, we get:
[tex]E_{cell}=2.76-\frac{0.059}{2}\times \log(\frac{1.59}{5.15})\\\\E_{cell}=2.74V[/tex]
Hence, the cell voltage of the given chemical reaction is 2.74 V
