Question 36 In a particular redox reaction, Fe2+ is oxidized to Fe3+ and Au3+ is reduced to Au. How many electrons are t...

Given,

In a particular Redox reaction Fe²⁺ is oxidized to Fe³⁺ and Au³⁺ is reduced to Au

Number of electrons are transferred in this reaction = ?

First we have to write the complete to Redox reaction:

In a particular Redox reaction Fe²⁺ is oxidized to Fe³⁺ and Au³⁺ is reduced to Au.

At anode: oxidation (loss of electrons)

Fe²⁺ ➝ Fe³⁺ + e^-

Multiplying the above equation by 3, we get

3 × [Fe²⁺ ➝ Fe³⁺ + e^-]

3Fe²⁺ ➝ 3Fe³⁺ + 3e^- ----------(I)

At cathode: reduction (gain of electrons)

Au³⁺ + 3e^- ➝ Au -------------(II)

Now we have to write the completed redox reaction

Adding equation (I) and (II), 

3Fe²⁺ ➝ 3Fe³⁺ + 3e^-

Au³⁺ + 3e^- ➝ Au

Adding both the above equation, we get

3Fe²⁺ + Au³⁺  ➝ 3Fe³⁺ + Au

So, 3 moles of electrons are transferred in this reaction

37)

Calculating the pH of a buffer solution = ?

Given,

[HC₂H₃O₂] = 0.317 M

[NaC₂H₃O₂] = 0.171 M

K_a for HC₂H₃O₂ = 1.81 × 10^-5

First we have to calculate pK_a for HC₂H₃O₂:

K_a for HC₂H₃O₂ = 1.81 × 10^-5

The formula used here is

pK_a = -logK_a

pK_a = -log(1.81 × 10^-5)

pK_a = 4.742 

Now we can calculate the pH of a buffer solution:

pH = pK_a + log[salt]/[acid]

pH = pK_a + log[NaC₂H₃O₂]/[HC₂H₃O₂] 

pH = 4.742 + log(0.171 M/0.317 M)

pH = 4.742 + (-0.268)

pH = 4.474

pH ≈ 4.47

(Rounding answer to 2 decimal places)

So, pH of a buffer solution is 4.47

Conclusion,

36) number of electrons transferred are = 6 moles

37) The pH of a buffer is 4.47 .