1.) In black-and-white film developing (an ancient technology for most of you),

Question

1.) In black-and-white film developing (an ancient technology for most of you), excess AgBr is removed from

the film negative by an aqueous solution of sodium thiosulfate (Na2S2O3; commonly called hypo for

hyposulfite), through formation of the complex ion Ag(S2O3)2^3-. The Kf of Ag(S2O3)2^3- is 4.7x10^13 and the Ksp of AgBr is 5.0x10^-13.

(c) How much (in g) of 1.0000000 g of AgBr does and how much does not dissolve in 1L of water and 1L of 1 M Na2S2O3?

2.)

How many moles/grams of oxygen and hydrogen are produced after 5 min of electrolysis at a current of 0.23 A? What is the volume of the gases at room temperature and atmospheric pressure (assume ideal gas law). What is the total number of electrons that passed the electrode/electrolyte interface?

3)

What is the current measured in a circuit involving the half cell reaction

Sn+4 + 2e- ---> Sn+2

which is occurring at a rate of 4.2 x 10-3 mole/hr?

4)

4) Calculate the potential in the following cell

Zn|Zn2+(1.00M)|| Cu2+(1.00M) |Cu

a) right at the beginning when you start using this voltaic cell,

b) when 50% of the Cu2+ ions have been consumed,

c) when 99% of the Cu2+ ions have been consumed, and

d) when 99.99% of the Cu2+ ions have been consumed.

Keep in mind that both, the Zn2+ and the Cu2+ concentrations are changing.

Explanation / Answer

2) , 2H+ + 2e- ---> H2 , O2- ------> O2 + 2e- ,

t= 5min = 5 x 60 = 300 s , i= 0.23 Amperes

moles = ( i x t /zF) , z = number of electrons involved in process ,

moles of O2 = moles of H2 = 0.23 x 300/96485x2) = 0.00035757

T = 298 K , P =1

V = nRT/P = 0.00035757 x 0.0821 x 298/1 = 0.00875 liters

charge = current x time = 0.23 x 300 = 69 coloumbs = N x 1.6 x 10^-19   ( electron charge = 1.6x10^-19 C , N = number of elctrons )

N = 4.3125 x 10^20 electrons

3) moles/time = ( i /zF) , z =2 ,

4.2 x 10^-3 = i /( 2 x 96485)

i = 810.47 amperes

4a) E cell = Eo(Cu2+/Cu) -Eo( Zn2+/Zn) - 0.059/2 log [Zn2+]/[Cu2+]

= -0.28 -(-0.76) -0.059/2 log ( 1/1)

= 0.48 volts

b) E cell = 0.48 -0.059/2 log ( 1/0.5)

= 0.47

c) E cell = 0.48 -0.059/2 log ( 1/0.1)

= 0.4505

d) 99.99 % of 1M = 0.9999 , Cu2+ left = 1-0.9999 = 0.0001

E cell = 0.48 -0.059/2 log ( 1/0.0001)

= 0.362

c) 1 gm AgBr = 1/187.77 = 0.0053257 moles , vol of water = 1L ,

Ksp =[Ag+][Br-] =s^2= 5 x10^-13 ,

s= 7.07 x 10^-7 moles/L = 7.07 x 10^-7 x 187.77 = 0.0001328 gm/L

hence amount dissolved = 0.0001328 gm and amount undissolved = 1-0.0001328 = 0.999867 gm

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