Question

If 6.539 × 10⁻² g of metallic zinc is added to 100 ml saturated solution of AgCl. Find the value of log₁₀$\frac{[Zn^{2+}]}{[Ag^+]^2}$.
How many moles of Ag will be precipitated in the above reaction. Given that
$Ag^+ + e^- \longrightarrow Ag; E^0 = 0.80V ;$
$Zn^{2+} + 2e^- \longrightarrow Zn; E^0 = -0.76V$
(It was given that Atomic mass of Zn = 65.39)

Answer 1

Answer:

10^-5 moles of Ag will be precipitated

Explanation:

Given:

Weight of metallic zinc = 6.539 × 10^-2 g

Volume of AgCl solution = 100ml

To find value of log [Zn2+]/[Ag2+]

For finding the solution, we need to write the cell reaction

Ag+1/2Cl2 ==> AgCl2

Ag===> Ag+ + e+

1/2Cl2 + e- ==> Cl-

Ag+ + Cl- ==> AgCl

The cell circuit will be given by

Ag|Ag+|AgCl||Cl-|Cl2Pt

Ag+(aq) + Cl- (aq) ==> AgCl (solid)

Nle we need to find Delta G

= -109-(-129) -77

= -57 KJ/mol

Delta G is also given by

Delta Gamma = - nFE°

Hence

E° = -57000/96500

= 0.59V

Another formula for Delta G is

= -2.303RTlogK

= -2. 303* 8.314* 298 log k

Thus

Log k = 57000/2.303*8.314*298

= 9.9

= 10

Hence Taking log on both sides

K= 10^10

We know

Ksp= 1/k

=1/10^10

= 10^-10

Log ksp= -10

2 Ag + Cl2 = 2 AgCl

Ksp of AgCl= [Ag][Cl]

= S*S

= S^2

We will use Nernst equation now

E cell = E° cell - 0.0591/n * log[Zn2+]/[Ag2+]

This

[Zn2+]/[Ag2+]= 1.57*2/0.059

= 52.8

Hence taking log on both sides

K= 10^52.8

Thus, 10^-5 moles of Ag will be precipitated.