To deposit 1 mole of aluminium from a solution of AlCl 3- the amount ofA- 0-33 FaradayB- 3 FaradayC- 1 FaradayD- 1-33 Faraday

The correct option is B 3 Faraday The amount of electricity required = 3 Faraday. ...

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Carrier Concentration

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To deposit 1 mole of aluminium from a solution of AlCl_3,the amount of

1.33 Faraday

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0.33 Faraday

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3 Faraday

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1 Faraday

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Amount of electricity that can deposit 108 gm of silver from AgNO₃ solution is

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Electricity passed through a solution of alcl3 ,13.5gm al is deposited no.of Faraday must be

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C u S O_{4}

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A) Electrolysis of aq. $N a_{2} S O_{4}$ using Pt electrodes	1) 1 Faraday
B) The charge carried by $6.023 \times 10^{23}$ electrons is	2) 3 Faradays
C) The amount of electricity required to deposit 27 grams of Aluminium at cathode from molten $A l_{2} O_{3}$ is	3) $H_{2 (g)} /$ Pt
D) A gas in contact with an inert electrode	4) $O_{2}$ at anode $H_{2}$ at cathode

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