1. Calculate the moalr solubility of AgBr in a 2.7 M NH3 solution = ____________

Question

1. Calculate the moalr solubility of AgBr in a 2.7 M NH3 solution

= ______________________M

Select all that apply

2. The molar solubility of AgCl in 6.5 x10^-3 M AgNO3 is 2.5x10^-8 M. In deriving Ksp from these data, which of the following assumptions is (are) reasonable

A. Ksp is the same as solubility

B. Ksp of AgCl is the same in 2.5x10^-8 M AgNO3 as in pure water

C. Solubility of AgCl is independent of the concentration of AgNO3

D. [Ag+] does not change significantly upon addition of AgCl to 6.5x10^-3 M AgNO3

E. [Ag+] after the addition of AgCl to 6.5x10^-3 M AgNO3 is the same as it would be in pure water

3. A 265 mL solution of NaCl was electrolyzed for 6.90 min. If the pH of the final solution was 12.02, calculate the average current used? = __________________A

4. What is the standard emf of a galvanic cell made of a Pb electrode in a 1.0 M Pb(NO3)2 solution and a Al electrode in a 1.0 M Al(NO3)3 solution at 25C?

Ecell= _____________________V

Explanation / Answer

#1

AgBr is a sparingly soluble salt. The ionization reaction for it can be written as

AgBr (s) <---------------> Ag+ (aq) + Br- (aq) ..............Ksp = 4.1 x 10^-13 ......... (I)

( Please note that this value is taken from online resources and can affect the final answer. Please check the value given to you and make the changes accordingly.)

Ag+ ions can react with NH3 to form a complex Ag(NH3)2^+ . The formation constant for this complex is 1.6 x 10^7 .

( Please note that this value is taken from online resources and can affect the final answer. Please check the value given to you and make the changes accordingly.)

The equation for this complex formation reaction can be written as

Ag+ (aq) + 2 NH3 (aq) <--------------> Ag(NH3)2^+ (aq) ...............Kf = 1.6 x 10^7 ............(II)

Let's add equations (I) and (II)

AgBr (s) <---------------> Ag+ (aq) + Br- (aq) ...................................Ksp = 4.1 x 10^-13

+ Ag+ (aq) + 2 NH3 (aq) <--------------> Ag(NH3)2^+ (aq) ...............Kf = 1.6 x 10^7

AgBr (s) + 2 NH3 (aq) <--------------> Ag(NH3)2^+ (aq) + Br- (aq) ..............K = 4.1 x 10^-13 * 1.6 x 10^7 = 6.56 x 10^-6

Expression for K can be written as

K = [ Ag(NH3)2^+ ] [ Br-] / [ NH3]^2

6.56 x 10^-6 = [ Ag(NH3)2^+ ] [ Br-] / [ NH3]^2

Let's draw ICE table for the above reaction.

AgBr (s) + 2 NH3 (aq) <--------------> Ag(NH3)2^+ (aq) + Br- (aq)

Let's plug in equilibrium values

6.56 x 10^-6 = (x) (x) / ( 2.7 - x)^2

6.56 x 10^-6 = x^2 / ( 2.7 - x)^2

Take square root on both sides

Sq rt ( 6.56 x 10^-6) = x / 2.7 - x

2.56 x 10^-3 = x / 2.7 - x

0.00256 ( 2.7 - x) = x

0.006915 - 0.00256x = x

x + 0.00256x = 0.006915

1.00256x = 0.006915

x = 0.006915 / 1.00256

x = 0.006898

x = 6.90 x 10^-3

The molar solubility of AgBr in 2.7 M NH3 solution is 6.90 x 10^-3 M

_______________________________________________________________________________________

#3

The species that will undergo reduction is water because reduction potential of water is much lower than Na+ ions.

Therefore it is easier to reduce water than Na+

The reaction can be written as

2 H2O (l) + 2 e- -------------> H2 (g) + 2 OH- (aq)   

The species that undergoes oxidation is Cl- . The reaction is given below.

2 Cl- (aq) -----------------> Cl2 (g) + 2e-

The overall reaction is

2 H2O (l) + 2 Cl- (aq) -------------> H2 (g) + Cl2 (g) + 2 OH- (aq)

The OH- ions are responsible for the pH of the solution.

Let's find conc. of OH- using the given pH

pOH = 14 - pH

pOH = 14 - 12.02

pOH = 1.98

pOH = - log [OH-]

1.98 = - log [OH-]

[OH-] = 10^-1.98

[OH-] = 0.01047 M
Using reduction reaction of water, let's find moles of electrons transferred.

0.01047 mol / L OH- * 265 mL * 1 L / 1000 mL * 2 mol e- / 2 mol OH- = 0.002775 mol e-

Charge carried by 1 mol electrons is 96500 coulombs

we have 0.002775 mol e-. Charge carried by them is

0.002775 mol e- * 96500 coulombs / 1 mol e- = 267.8 coulombs.

Current can be calculated as A = coulombs/ time in seconds

We have passed the current for 6.90 minutes. Let's convert that to seconds.

6.90 minutes * 60 seconds / 1 min = 414 seconds.

Current = 267.8 coulombs / 414 seconds

Current = 0.647 Ampere

The average current was 0.647 ampere

_______________________________________________________________________________________

#4

Using standard reduction potential table, let's find the reduction potentials of Pb and Al

Pb 2+ (aq) + 2e- ----------> Pb (s) ........... E^0 = -0.13 V

Al^3+ (aq) + 3e- -------------> Al(s) ...............E^0 = -1.66 V

Since reduction potential for Pb is higher, it will behave as cathode and Al will behave as anode.

Reaction at cathode = Pb 2+ (aq) + 2e- ----------> Pb (s)

Reaction at anode = Al(s) ------------> Al^3+ (aq) + 3e-

E^0 cell = E^0 cathode - E^0 anode

E^0 cell = -0.13 V - ( -1.66 )

E^0 cell = 1.53 V

The overall cell reaction can be written as

Multiply cathode reaction by 3 and anode reaction by 2 and add them

3 Pb 2+ (aq) + 6e- ----------> 3 Pb (s)

+  2 Al(s) ------------> 2 Al^3+ (aq) + 6 e-

_____________________________________

3 Pb^2+ (aq) + 2 Al(s) -------------> 3 Pb (s) + 2 Al^3+ (aq)

reaction quotient for this reaction can be written as

Q = [ Al3+] ^2 / [Pb2+] ^3

Substituting the given concentrations, we get

Q = (1)^2 / (1)^3 =1

The emf of the cell can be found out using Nernst equation.

E cell = E^0 cell - RT/nF * ln Q

Here, R = 8.314 J/ mol K

T = 298 K

F = 96500 C

n = 6 e-

Let's plug in above values.

E cell = 1.53 V - 8.314 J/ mol K * 298 K / 6 e- * 96500 C * ln (1)

E cell = 1.53 V - 0

E cell = 1.53 V

_____________________________________________________________________________________________

AgBr NH3 Ag(NH3)2^+ Br- I - 2.7 0 0 C - -x +x +x E - 2.7 - x x x

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