Answer: The standard potential for this cell is +1.49 V at 25C.
Explanation:
[tex]E^0_{[Sn^{2+}/Sn]}=-0.14V[/tex]
[tex]E^0_{[Ti^{2+}/Ti]}=-1.63V[/tex]
As titanium has lower reduction potential, it will act as anode and tin will acts as cathode.
[tex]Ti+Sn^{2+}\rightarrow Ti^{2+}+Sn[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ti^{2+}]}{[Sn^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]25^oC=273+25=298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^0=E^0_{cathode}- E^0_{anode}=-0.14-(-1.63)=1.49V[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 1.49 V
[tex]E_{cell}[/tex] = emf of the cell = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=1.49-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}[/tex]
[tex]E_{cell}=1.49V[/tex]
The voltage of the voltaic cell is 1.49 V.
A voltaic cell is a cell in which electrical energy is produced by a spontaneous chemical reaction.
The equation of the cell is; Ti(s) + Si^2+(aq) -----> Ti^2+(aq) + Si(s)
E°cathode = -0.14 V
E°anode = -1.63 V
E°cell = (-0.14 V) - (-1.63 V) = 1.49 V
Using the Nernst equation;
E = E°cell - 0.0592/n log[Ti^2+]/[Si^2+]
E = 1.49 V - 0.0592/2 log(1)/(1)
E = 1.49 V
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