Question

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction isAgCl (s) + e- ? Ag (s) + Cl-(aq) E° = +0.222 VThe concentrations of chloride ion in the two compartments are 0.0156 M and 1.55 M, respectively. The cell emf is __________ V.a 0.232b 22.1c 0.212d 0.00223e 0.118

Answer 1

Answer : The cell emf for this cell is 0.118 V

Solution :

The balanced cell reaction will be,  

$AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq)$

Oxidation half reaction (anode):  $Ag+Cl^-\rightarrow AgCl$

Reduction half reaction (cathode):  $AgCl\rightarrow Ag+Cl^-$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^-{cathode}}{[Cl^-{anode}]}$

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^-{diluted}}{[Cl^-{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^-{diluted}]$ = 0.0156 M

$[Cl^-{concentrated}]$ = 1.55 M

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

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0156}{1.55}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V