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 :

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

where,

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

$E_{cell}$ = ?

$[Cl^{-}{diluted}]$ = 0.0222 M

$[Cl^{-}{concentrated}]$ = 2.22 M

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

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V