Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(P_H2)= 0.87atm.

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) > Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M =  0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 =  0.76 - 0.0592/2logQ

0.660 - 0.76  =  - 0.0592/2logQ

-0.1 =  - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 =  [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92  × 10^-3/[H^+]^2

[H^+]^2 = 7.92  × 10^-3/2.39 × 10^3

[H^+] = 1.82  × 10^-3

pH = -log[H^+]

pH = -log[ 1.82  × 10^-3]

pH = 2.7

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given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g