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

6 Transcribed: Question 36 In a particular redox reaction, Fe2+ is oxidized to Fe3+ and Au3+ is reduced to Au. How many electrons are transferred in this reaction? (Hint: Balance the reaction) Question 37 Calculate the pH of a buffer that is 0.317 M HC2H3O2 and 0.171 M NaC2H3O2. The Ka for HC2H302 is 1.81 x 10-5 Round answer to 2 decimal places.

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

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

Ka for HC2H3O2 = 1.81 × 10-5

First we have to calculate pKa for HC2H3O2:

Ka for HC2H3O2 = 1.81 × 10-5

The formula used here is

pKa = -logKa

pKa = -log(1.81 × 10-5)

pKa = 4.742

Now we can calculate the pH of a buffer solution:

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

pH = pKa + 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 .

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