查看完整版本 : electrochemistry

tadaima789 2012-4-29 09:37 PM

electrochemistry

[font=cd9800338d999ea0101be1a8#700c00][size=3]2. A current is passed through a 500.0ml solution of CaI2. The following electrode reactions occur:[/size][/font]
[font=cd9800338d999ea0101be1a8#700c00][size=3]
[/size][/font]
[font=cd9800338d999ea0101be1a8#700c00][size=3]anode : 2I- -> I2 +2e[/size][/font]
[font=cd9800338d999ea0101be1a8#700c00][size=3]cathode : 2H2O+2e -> H2 + 2OH-[/size][/font]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]
After some time, analysis of
the solution shows that41.5 mmol of I[/size][/font][font=cd9800338d999ea0101be1a8#700c00][size=3]2[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]has been formed.[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]
(a) How many faradays of charge have passed through the solution?[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]
(b) How many coulombs?(c) What volume of dry H[/size][/font][font=cd9800338d999ea0101be1a8#700c00][size=3]2[/size][/font][font=cd9800338d999ea0101be1a8#700c00][size=3]at STP has been formed?[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]
d) What is the pH
of the solution?[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]
[/size][/font] [/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]是課本後的習題,只有答案沒有解釋,part A不明白 faradays 到底要怎算? [/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]part B coulombs不是要時間計算嗎?但題目沒說明[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]part C是coulombsX 1/ 9.65X10^4X 1/4 X STP of H2?   STP是可以查表查到定自己計?[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]part D pH of solution要計返Q再計H+之後pH=-log[H+]去計???[/size][/font][/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]
[/size][/font] [/size]
[size=73px][font=cd9800338d999ea0101be1a8#700c00][size=3]請幫忙~謝謝[/size][/font][/size]

jmlo 2012-4-30 06:03 AM

回覆 1# 的帖子

[quote] part A不明白 faradays 到底要怎算? [/quote]
Note the conversion of these units.
1 Faraday = charges carried by 1 mole of electrons

[quote] part B coulombs不是要時間計算嗎?但題目沒說明 [/quote]
Once you know the answer above, you can solve this question by using the following conversion
1 Faraday = 96485.3399 Coulomb

[quote] STP是可以查表查到定自己計? [/quote]
You can check the molar volume of gas at STP from wikipedia or other reference books.

[quote] part D pH of solution要計返Q再計H+之後pH=-log[H+]去計??? [/quote]
Once you know how many moles of H2 is formed, you can work out the amount of OH-. Since pH + pOH = 14, you can find out the pH of the solution accordingly.

tadaima789 2012-4-30 01:39 PM

A)是否mmol轉g
41.5X(1/10^3)X126.9X0.5=2.633 g
1g I2requires = 4/126.9    F charge
2.633 g I2 requires = (4X2.633) / 126.9 = 0.083 F

我的答案計極都同書有小小出入...係原子量唔同定我分條計錯:smile_27:

jmlo 2012-4-30 02:39 PM

回覆 3# 的帖子

What's the answer for part A?

You can solve part A in the following way:

Note that each mole of I2 formed gives up 2 moles of electrons. It means that 41.5 mmol of I2 in total provides 41.5*2 = 83 mmol of electrons. The charges passed through the solution is thus 83*10^-3*1 = 0.083 Faradays.
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