Question

Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should A) increase by 0.06 V. D) decrease by 0.18 V. B) increase by 0.18 V. E) increase by 0.35 V. C) decrease by 0.06 V.

Answer 1

B) increase by 0.18 V

Step-by-step explanation:

The given chemical spontaneous reaction is :

`2 AgCl_((s)) + Zn _((s)) \to 2Ag (s) + 2Cl^- _((aq)) + Zn ^(2+) _((aq))`

By applying Nernst Equation:

`E_(cell) = E^0 - (0.059)/(n) log [(product)/(reactant)]`

here ;

n = number of electrons transferred in the reaction

n =2

`E^0 = E^0_(cathode ) - E^0_(anode)`

`E^0 = E^0_(Ag^+/Ag ) - E^0_(Zn^(2+)/Zn)`

`E^0 =+0.80 \ V -(-0.76 \ V)`

`E^0 =1.56 \ V`

When it happens to occur that the concentration of chlorine (aq) and Zn²⁺ (aq) is 1 M ;

`E^0_(cell) = E^0` is as follows:

`E_(cell) = E^0 - (0.059)/(n) log [(product)/(reactant)]`

`E_(cell) = 1.56 - (0.059)/(n) log [([Zn^(2+)])/([Cl^-]^2)]`

`E_(cell) = 1.56 - (0.059)/(n) log (1) \ \ \ \ \ \ \ ( where \ log (1) = 0)`

`E_(cell) = 1.56 \ V`

Now; the
`E_(cell)` value in the decreased concentration of chlorine (aq) ion is calculated as:

`E_(cell) = E^0 - (0.059)/(n) log [(product)/(reactant)]`

`E_(cell) = 1.56 - (0.059)/(n) log [([Zn^(2+)])/([Cl^-]^2)]`

`E_(cell) = 1.56 - (0.059)/(n) log (1*0.001^2)`

`E_(cell) = +1.737 \ V`

Hence; the change in voltage =
`E_(cell) - E^0`

= 1.737 - 1.56

= 0.177 V

≅ 0.18 V

We therefore conclude that: since the
`E^0_(cell)` value after the decreased concentration of Chlorine is greater than the
`E^0` before the change; then there is increase in the value by 0.18 V