1.50 g chromium metal from Cr^3+ solution in 30.0 s

1.50 g chromium metal from Cr^3+ solution in 30.0 s

 

 

What current (in A) is needed to plate out

1.50 g of chromium metal from Cr^3+ solution in 30.0 seconds?

 

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▶ 참고: 전기분해의 화학량론

[ https://ywpop.tistory.com/4461 ]

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Cr의 몰질량 = 52.00 g/mol 이므로,

1.50 g / (52.00 g/mol) = 0.028846 mol Cr

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

Cr^3+ + 3e^- → Cr

---> n = 3

 

 

 

(0.028846 mol) × 3 = 0.086538 mol e^-

 

 

 

(0.086538 mol) × (96500 C / mol) = 8350.917 C

 

 

 

Q = I t

 

I = Q / t

= 8350.917 C / 30.0 s

= 278.3639 A

 

 

 

한번에 계산하면,

전류(A) = 질량(g) / 몰질량 × 이동한 전자 × 96500 / 시간(s)

= (1.50) / (52.00) × (3) × (96500) / (30.0)

= 278 A

 

 

 

답: 278 A

 

 

 

 

[키워드] What current is needed to deposit 1.50 g of chromium metal from a solution of Cr^3+ in a period of 30.0 seconds?