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)-(0.0592)/(n)\log \frac{[Cl^-{cathode}}{[Cl^-{anode}]}`

`E_(cell)=E^o_(cell)-(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-(0.0592)/(1)\log (0.0156)/(1.55)`

`E_(cell)=0.118V`

Therefore, the cell emf for this cell is 0.118 V