Asked Nov 11, 2019

In the electrolysis of water, how long will it take to produce 1.000 × 102 L of H2 at STP (273 K and 1.00 bar) using an electrolytic cell through which a current of 58.50 mA flows? (Answer in Years please!)


Expert Answer

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Step 1

Oxidation is the process that involved the loss of electrons and oxidized to cation.  Hence when a molecule loses an electron, that molecule is said to be oxidized.  Metals have tendency to lose electrons and form cation such as sodium oxidized to sodium ion.

All metals are arranged in the metal activity series according to their reactivity. The more reactive metal is placed above in the series and will be good reducing agent whereas less reactive metal is placed below in the series.

Step 2

According to Faraday's law of electrolysis:


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Exсxt W = F W-Massin g of an ion discharged at any electrode E Equivalent mass of that ion. c Current Passed in Ampere t=time (in sec) during which electricity is passed F= 1 faraday 96500C.

Step 3

Current = 58.50 mA = 58.50 x 10-3A

Volume of hydrogen gas = 1.000 x 102 L

Temperature = 273 K

Pressure = 1.00 bar = 0.987 atm

Calculate mole of hydrogen gas:


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PxV=nxRxT 0.987atm 1.000x 10L= n x 0.083 14L .atm/K.mol x 273K 0.987atmx1.000x10? L 0.08314L.atm/K.mol x 273K n 4.35 moles


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