Electrolysis and Faraday's Law · 电解和法拉第定律
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Faraday's constant/ˈfærədeɪz ˈkɒnstənt/ | 法拉第常数 | fǎ lā dì cháng shù |
| electrolysis/ɪlekˈtrɒləsɪs/ | 电解 | diàn jiě |
Counting atoms with a stopwatch
- Pass current through a solution and metal plates out.
- The longer and stronger the current, the more metal forms.
- Charge is just electrons, and electrons build atoms.
- So a clock and an ammeter can count atoms.
用秒表数原子
- 让电流通过溶液,金属就镀出来。
- 电流越久越强,生成的金属越多。
- 电荷不过是电子,而电子构建原子。
- 所以一个钟表和一个电流表就能数原子。
Charge equals current times time
- The total charge is $Q = It$, current times time.
- More current or more time means more charge.
- That charge is a flood of electrons.
电荷等于电流乘时间
- 总电荷是 $Q = It$,即电流乘时间。
- 更大的电流或更长的时间意味着更多电荷。
- 那份电荷是一股电子的洪流。
A current of $5\ \text{A}$ flows for $200\ \text{s}$. The total charge (in C)? · 电流$5\ \text{A}$流过时间$200\ \text{s}$。总电荷（库仑）是多少？
$Q = It = 5 \times 200 = 1000\ \text{C}$.
Faraday's constant
- Divide the charge by Faraday's constant 法拉第常数 $F$ to get moles of electrons.
- $F \approx 96{,}500\ \text{C/mol}$.
- The moles of electrons drive the electrode reaction.
法拉第常数
- 用电荷除以法拉第常数 $F$ 得到电子的摩尔数。
- $F \approx 96{,}500\ \text{C/mol}$。
- 电子的摩尔数驱动电极反应。
To convert charge into moles of electrons, you divide by... · 要将电荷转换为电子摩尔数，需除以...
Moles of electrons $= Q / F$. · 电子摩尔数$= Q / F$。
From electrons to product
- The half-reaction tells how many electrons each atom needs.
- Divide the moles of electrons by that number to get moles of product.
- More charge means more metal deposited.
从电子到产物
- 半反应告诉你每个原子需要多少电子。
- 用电子的摩尔数除以那个数目,得到产物的摩尔数。
- 电荷越多,镀出的金属越多。
Run an electrolysis cell · 运行电解池
Push a current through an electrolyte and see which ions are discharged at each electrode. · 向电解质通入电流，观察每个电极上哪些离子被放电。
Passing more charge through the cell deposits... · 通过电池的电荷越多，沉积的...
More charge means more electrons and more product. · 更多电荷意味着更多电子和更多产物。
Depositing copper ($\text{Cu}^{2+} + 2e^- \to \text{Cu}$) needs 2 electrons per atom. For $2\ \text{mol}$ of electrons, how much copper?
- $2\ \text{mol e}^- \div 2 = 1\ \text{mol Cu}$.
- So $1$ mole of copper plates out.
镀铜($\text{Cu}^{2+} + 2e^- \to \text{Cu}$)每个原子需要 2 个电子。对 $2\ \text{mol}$ 电子,能得多少铜?
- $2\ \text{mol e}^- \div 2 = 1\ \text{mol Cu}$。
- 所以镀出 $1$ 摩尔铜。
$\text{Cu}^{2+} + 2e^- \to \text{Cu}$ needs 2 electrons per atom. How many moles of Cu from $4\ \text{mol}$ of electrons? · $\text{Cu}^{2+} + 2e^- \to \text{Cu}$每原子需要2个电子。从$4\ \text{mol}$个电子可以得到多少摩尔Cu？
$4\ \text{mol e}^- \div 2 = 2\ \text{mol Cu}$.
Order the steps of a Faraday's-law calculation. · 排列法拉第定律计算的步骤顺序。
Charge, then moles of electrons, then moles of product. · 电荷，然后电子摩尔数，然后产物摩尔数。
Depositing silver ($\text{Ag}^+ + e^- \to \text{Ag}$) needs ____ electron(s) per atom. · 沉积银（$\text{Ag}^+ + e^- \to \text{Ag}$）每原子需要____个电子。
$\text{Ag}^+$ gains just one electron per atom. · $\text{Ag}^+$每原子只获得一个电子。
Convert charge to moles of electrons with Faraday's constant first, then use the half-reaction's electron count to get moles of product. Watch the electrons per atom ($\text{Cu}^{2+}$ needs 2, $\text{Ag}^+$ needs 1). And $Q = It$ needs current in amperes and time in seconds.
先用法拉第常数把电荷换算成电子的摩尔数,再用半反应的电子数目得到产物的摩尔数。注意每个原子的电子数($\text{Cu}^{2+}$ 需要 2 个,$\text{Ag}^+$ 需要 1 个)。而且 $Q = It$ 需要电流用安培、时间用秒。
In electrolysis 电解, the charge $Q = It$ delivers electrons. Divide by Faraday's constant ($\approx 96{,}500\ \text{C/mol}$) for moles of electrons, then divide by the half-reaction's electrons-per-atom for moles of product. More charge deposits more metal.
在电解中,电荷 $Q = It$ 传递电子。除以法拉第常数($\approx 96{,}500\ \text{C/mol}$)得到电子的摩尔数,再除以半反应每原子的电子数,得到产物的摩尔数。电荷越多,镀出的金属越多。