Question

A galvanic (voltaic) cell consists of an electrode composed of titanium in a 1.0 M titanium(II) ion solution and a second electrode composed of tin in a 1.0 M tin(II) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25C.

Answer 1

Answer: The standard potential for this cell is +1.49 V at 25C.

Explanation:

$E^0_{[Sn^{2+}/Sn]}=-0.14V$

$E^0_{[Ti^{2+}/Ti]}=-1.63V$

As titanium has lower reduction potential, it will act as anode and tin will acts as cathode.

$Ti+Sn^{2+}\rightarrow Ti^{2+}+Sn$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ti^{2+}]}{[Sn^{2+}]}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $25^oC=273+25=298K$

n = number of electrons in oxidation-reduction reaction = 2

$E^0=E^0_{cathode}- E^0_{anode}=-0.14-(-1.63)=1.49V$

Where both $E^0$ are standard reduction potentials.

$E^o_{cell}$ = standard electrode potential of the cell = 1.49 V

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.49-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}$

$E_{cell}=1.49V$