"250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be [ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]"

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Accepted Answer

"Millimoles of Au⁺ = 0.1 × 250 = 25

Mole of Au⁺ = ${{25} \over {1000}}$ = ${1 \over {40}}$

Charge passed = I × t = 1 × 15 × 60 = 900 C

moles of e^– passed = ${{900} \over {96500}}$ = ${9 \over {965}}$

Only gold will be deposited as quantity of charge passed is less than the amount of Au⁺ present."

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO₃ and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be

[ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]

Solution

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250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be [ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]

Solution

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250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO₃ and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be

[ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]