Question

Find out emf of cell Zn / Zn2+ // Cu (1M) / Cu E° For Zn2+ = -0.76 v , E° For Cu2+ = 0.34

Answer 1

Emf of cell= 0.34-(-0.76) =0.34+0.76=1.1

It is the standard electrode potential cor the given arrangement.

Answer 2

Emf of the cell is +1.10 V

Explanation:

$Zn+Cu^{2+}(1M)\rightarrow Cu+Zn^{2+}(1M)$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{1}{[X]}$

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

$E^o_{cell}$ = standard electrode potential of the cell =  $E^o_c-E^0_a=E^o_{Cu}-E^0_{Zn}=0.34-(-0.76)=1.10V$

$E_{cell}$ = emf of the cell

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

$E_cell=1.10-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}$

$E_{cell}=1.10V$

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