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Electrolysis of a solution of HSO(4)^(-)...

Electrolysis of a solution of `HSO_(4)^(-)` ions produces `S_(2)O_(8)^(-)`. Assuming `75%` current efficiency, what current should be employed to achieve a production rate of 1 mole of `S_(2)O_(8)^(-)` per hour?

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The correct Answer is:
`+71.5amp`
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Electrolysis of a solution of HSO_(4)^(-1) ions produces S_(2)O_(8)^(2-) . Assuming 75% current efficiency, what current should be employed to achieve a production rate of 1 "mole" of S_(2)O_(8)^(2-) per hour?

What is the average oxidation state of oxygen In S_2O_8^(2-)

The number of gm atoms of oxygen present in 0.2 mole of H_(2)S_(2)O_(8) is.

During the electrolysis of conc H_(2)SO_(4) , it was found that H_(2)S_(2)O_(8) and O_(2) liberated in a molar ratio of 3:1 . How many moles of H_(2) were found of moles of H_(2)S_(2)O_(8) ? ( Express your answer as :3xx mol e s of H_(2), integer answer is between 0 and 5 0

What is the product formed when: Na_(2)S_(2)O_(4) is hydrolysed

Electrolysis of aq. Solution of CH_(3)COOK is caried out. If 2 amp current is passed for 402 minutes. What volume (at STP) of gases could be produced if current efficiency is 80% (assume no absorption of gases)

In this reaction: S_(2)O_(8)^(2-)+2I^(-) to 2SO_(4)^(2-)+I_(2)

Hydrogen peroxide can be prepared by successive reaction: 2NH_(4)HSO_(4) rarr H_(2)+(NH_(4))_(2)S_(2)O_(8) (NH_(4))_(2)S_(2)O_(8)+2H_(2)O rarr 2NH_(4)HSO_(4)+H_(2)O_(2) The first reaction is an electrolytic reaction the second is steam distillation. what amount ofcurrent would have to be used in first reaction to produce enough intermediate to yield 100g pure H_(2)O_(2) per hour ? Assume 50% anode current efficiency.

Hydrogen peroxide can be prepared by successive reaction: 2NH_(4)HSO_(4) rarr H_(2)+(NH_(4))_(2)S_(2)O_(8) (NH_(4))_(2)S_(2)O_(8)+2H_(2)O rarr 2NH_(4)HSO_(4)+H_(2)O_(2) The first reaction is an electrolytic reaction the second is steam distillation. what amount ofcurrent would have to be used in first reaction to produce enough intermediate to yield 100g pure H_(2)O_(2) per hour ? Assume 50% anode current efficiency.

H_(2)O_(2) can be produced by the Ammonium hydrogen sulphate. Reactions occuring in electrolytic cell. NH_(4)HSO_(4) rarr NH_(4)^(+) +H^(+) +SO_(4)^(2-) 2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-) (Anode) 2H^(+) +2e^(-) rarr H_(2) (cathode) (NH_(4))_(2)S_(2)O_(8) +2H_(2)O rarr 2NH_(4)HSO_(4) +H_(2)O_(2) Hydrolysis of Ammonium persulphate Assume 100% yield of hydrolysis reaction. What is the current efficiency when 100 Amp current is passed for 965 sec, in order to produce 17 gm of H_(2)O_(2) .

RESONANCE ENGLISH-ELECTROCHEMISRY-Exercise
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