Question

a galvonic cell is formed by the combination of zn rid immeresed in zinc chloride solution of concentration 0.5 molar and li rod immersed in lithium chloride solution of concentration 0.15 molar. Give the cell representaion, cell reaction and calculate emf of the cell at 25 degree Celcius. The standard electrod potential of li and zn electrodes are -3.04V and -0.76V respectively.​

Answer 1

Answer: a) $Li/Li^{+}(0.15M)//Zn^{2+}(0.5M)/Zn$

b) $2Li+Zn^{2+}(0.5)\rightarrow 2Li^{+}(0.15)+Zn$

3. 2.24 V

Explanation:

Here Li undergoes oxidation by loss of electrons, thus act as anode. Zinc undergoes reduction by gain of electrons and thus act as cathode.

a) The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

$Li/Li^{+}(0.15M)//Zn^{2+}(0.5M)/Zn$

2. The cell reaction will be represented as:

$2Li+Zn^{2+}(0.5)\rightarrow 2Li^{+}(0.15)+Zn$

3. $E^0=E^0_{cathode}- E^0_{anode}$

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

$E^0_{[Li^{+}/Li]}= -3.04V$

$E^0_{[Zn^{2+}/Z]}=-0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Li^{+}/Li]}$

$E^0=-0.76- (-3.04V)=2.28V$

Using Nernst equation:

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

where,

The $E^0$ values have to be reduction potentials.  

n = number of electrons in oxidation-reduction reaction

F = faradays constant = 96500 C

R= gas constant = 8.314 J/Kmol

T = temperature =$25^0c=25=273=298K$

Putting in the values:

$E_cell=2.28-\frac{2.303\times 8.314\times 298}{2\times 96500}\log \frac{[0.15]^2}{[0.5]}$

$E_cell=2.24V$

Thus emf of the cell at 25 degree Celcius is 2.24 V