Question

Calculate the emf of the following cell at 25°C. 
Ag(s)/Ag+ (10^-3)// Cu^2+ (10^-1m)/Cu(s) 
Given E°cell =+0.46V,and log 10^n=n.

Answer 1

Answer : The EMF of the cell at $25^oC$ is, 0.61 V

Solution :

The balanced cell reaction will be,

$2Ag(s)+Cu^{2+}(aq)\rightarrow 2Ag^+(aq)+Cu(s)$

Here silver (Ag) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ag^{+}]^2}{[Cu^{2+}]}$

where,

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

$E^o_{cell}$ = standard electrode potential = 0.46 V

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

$E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(10^{-3})^2}{10^{-1}}=0.61V$

Therefore, the EMF of the cell at $25^oC$ is, 0.61 V

Answer 2

Answer:

0.314V

may be correct support