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.

B) increase by 0.18 V

Explanation:

The given chemical spontaneous reaction is :

By applying Nernst Equation:

here ;

n = number of electrons transferred in the reaction

n =2

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

is  as follows:

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

Hence; the change in voltage =

= 1.737 - 1.56

= 0.177 V

≅ 0.18 V

We therefore conclude that: since the value after the decreased concentration of Chlorine is greater than the before the change; then there is increase in the value by 0.18 V

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally