Question

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl− (aq) E° = +0.222 V The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M, respectively. The cell emf is ________ V. A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is (s) + (s) + (aq) E° = +0.222 V The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M, respectively. The cell emf is ________ V. 0.00222 0.232 0.118 22.2 0.212

Answer 1

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

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

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^(-){diluted}]}{[Cl^(-){concentrated}]}`

where,

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

`E_(cell)` = ?

`[Cl^(-){diluted}]` = 0.0222 M

`[Cl^(-){concentrated}]` = 2.22 M

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

`E_(cell)=0-(0.0592)/(1)\log (0.0222M)/(2.22M)`

`E_(cell)=0.118V`

Therefore, the cell emf for this cell is 0.118 V