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  • Need clarity, kindly explain! - Redox Reaction and Electrochemistry - JEE Main-3

The amount of lactic acid, HC3H5O3, production is a sample of muscle tissue was analysed by reaction with hydroxide ion. Hydrooxide ion was produced in the sample mixture by elctrolysis. The cathode reaction was 

2H_2O + 2e^- \Rightarrow H_{2(g)} + 2OH^{-}_{(aq)}

Hydroxide ion reacts with lactic acid as soon as it is produced. The end point of the reaction is detected with an acid -base indicator. It is required 115 secs for a currecnt of 15.6 mA to reach end point. How many grams of lactic acid  (a monoprotic acid) were present in the sample ?

  • Option 1)

    1.848 \times 10^{-3} g

  • Option 2)

    1.674 \times 10^{-3} g

  • Option 3)

    1.426 \times 10^{-3} g

  • Option 4)

    1.943 \times 10^{-3} g

 

Answers (1)

best_answer

As we have learnt,

 

Faraday's first law of electrolysis -

The mass of any substance deposited or dissolved at any electrode during electrolysis is directly propotional to the quantity of electricity passed through the solution.

- wherein

W\alpha\ Q

W = Z I t

Z = mass of electrolyte get deposited

Z = electro chemical equivalent

 

 No. of mole sof lactic acid = moles of H+ ised = faraday used in electricity

No. of faraday used  = =\frac{I\times t}{96500} = \frac{15.6\times 10^{-3}\times 115}{96500} = 1.86\times 10^{-5}

Mass of lactic acid = no of moles x molecular mass = 1.86\times 10^{-5} \times 90 = 1.674\times 10^{-3}

 


Option 1)

1.848 \times 10^{-3} g

Option 2)

1.674 \times 10^{-3} g

Option 3)

1.426 \times 10^{-3} g

Option 4)

1.943 \times 10^{-3} g

Posted by

prateek

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