Electrochemistry (A2)
A-Level Chemistry Topic 24 7:53 English narration · English + 中文 subtitles burned in
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Transcript
Two things in your pocket run on the same chemistry.
你口袋里的两样东西，靠的是同一种化学。
A battery: a reaction that wants to happen pushes electrons through a wire.
电池：一个本来就想发生的反应， 把电子推过导线。
And the charger: we push those electrons back, forcing that reaction into reverse — electrolysis.
而充电器：我们把电子反着推回去，强迫这个反应倒着进行， 这就是电解。
Today you learn the numbers behind both.
今天你要学的，就是这两件事背后的数字。
Welcome to A2 electrochemistry.
欢迎来到 A2 电化学。
First, electrolysis: what appears at each electrode, and how much.
先看电解：每个电极上会析出什么，析出多少。
Then electrode potentials: how we measure them, combine them, and use them to predict a reaction.
再看电极电势：我们如何测量它、如何把它们组合起来， 以及如何用它预测一个反应。
A quick reminder.
先快速回顾一下。
In electrolysis a direct current passes through a liquid full of free ions.
在电解中，直流电通过含有自由离子的液体。
Positive ions go to the cathode and gain electrons; negative ions go to the anode and lose them.
正离子移向阴极，在那里得到电子；负离子移向阳极，在那里失去电子。
Now the A2 question: what comes out, and how much?
而 A2 要问的是：到底析出什么？ 析出多少？
What is liberated at each electrode?
每个电极上会析出什么？
Three things decide it.
有三个因素决定。
First, the state of the electrolyte.
第一，电解质的状态。
A molten salt has only two kinds of ion: metal at the cathode, non-metal at the anode.
熔融的盐里只有两种离子：阴极得到金属，阳极得到非金属。
An aqueous solution also has water, so a reactive metal loses and hydrogen forms instead.
水溶液里还有水，所以活泼金属会输，析出的是氢气。
At the anode you usually get oxygen — but concentration counts too: a concentrated halide gives the halogen.
在阳极通常得到氧气——但浓度同样重要：浓的卤化物溶液会析出卤素单质。
Now the numbers.
现在来看数字。
The charge on one mole of electrons is the Faraday constant, ninety six thousand five hundred coulombs per mole.
一摩尔电子所带的电荷量叫做法拉第常量，约为九万六千五百库仑每摩尔。
Every calculation is then three steps.
此后每一道计算都是同样的三步。
One: the charge passed is current times time.
第一步：通过的电量等于电流乘以时间。
Two: divide that charge by the Faraday constant for moles of electrons.
第二步：用这个电量除以法拉第常量，得到电子的物质的量。
Three: use the half-equation to reach moles of product.
第三步：用半反应方程式换算出产物的物质的量。
And remember: that time is in seconds.
另外记住：时间要用秒。
A real one.
来一道真题。
A molten cobalt salt is electrolysed at zero point five amps, and zero point five four seven grams of cobalt forms at the cathode.
用零点五安的电流电解熔融的钴盐，阴极上析出零点五四七克钴。
How long did the current flow, in minutes?
电流通了多长时间，用分钟表示？
Step one: moles of cobalt, mass over relative atomic mass — zero point zero zero nine two nine moles.
第一步：钴的物质的量，用质量除以相对原子质量， 得到零点零零九二九摩尔。
Step two: each cobalt ion needs two electrons, so the charge is that, times two, times the Faraday constant: one thousand seven hundred and ninety two coulombs.
第二步：每个钴离子需要两个电子， 所以电量等于这个数乘以二，再乘以法拉第常量，约为一千七百九十二库仑。
Step three: charge over current is three thousand five hundred and eighty four seconds.
第三步：电量除以电流，是三千五百八十四秒。
Divide by sixty: fifty nine point seven minutes.
再除以六十：五十九点七分钟。
Notice the trap — the formula wants seconds.
注意这个陷阱：公式里用的是秒。
Electrolysis can even count the particles in a mole.
电解甚至能数出一摩尔里有多少个微粒。
Pass a steady current for a measured time, so you know the charge.
通入稳定的电流并记下时间，这样就知道了电量。
Weigh the cathode before and after for the mass deposited, and the half-equation gives moles of electrons.
电解前后各称一次阴极，得到析出的质量，再由半反应方程式得到电子的物质的量。
Divide, and out comes the Faraday constant.
两者相除，得到的就是法拉第常量。
Divide that by the charge on one electron, for the Avogadro constant.
再用它除以一个电子的电荷， 就得到阿伏伽德罗常量。
Now the second half.
现在进入下半部分。
Every half-cell has a tendency to be reduced, measured as a voltage.
每个半电池都有被还原的倾向，我们用电压来衡量它。
But a voltage is a difference, so we need a zero: the standard hydrogen electrode, hydrogen gas over platinum in acid, defined as exactly zero volts.
可是电压总是一个差值，所以我们需要一个零点：标准氢电极—— 氢气通过铂电极，浸在酸中，规定它正好是零伏。
Connect a half-cell to it under standard conditions and the reading is its standard electrode potential, always written as a reduction.
在标准条件下把半电池与它相连， 读数就是它的标准电极电势，而且总是写成还原的形式。
How is a half-cell built?
半电池到底怎么搭？
If the element is a metal, dip a strip of it into a solution of its own ions.
如果这种元素是金属，就把它的一小条浸入含它自身离子的溶液中。
But when both species are ions of the same element, like iron three and iron two, there is no solid to dip in.
但如果两种物种是同一元素的不同价态离子，比如三价铁和二价铁，就没有固体可以浸了。
So use a platinum electrode: it carries electrons without reacting.
这时改用铂电极：它只传导电子而不参与反应。
Join the halves with a salt bridge and read the voltmeter.
用盐桥把两个半电池连起来， 再读电压表。
Now the arithmetic.
现在来算。
The standard cell potential is the more positive electrode potential minus the less positive one.
标准电池电势等于较正的电极电势减去较不正的那个。
For the Daniell cell, copper is plus zero point three four and zinc is minus zero point seven six.
以丹尼尔电池为例，铜是正零点三四，锌是负零点七六。
Subtract: one point one zero volts, always positive in that order.
相减，得到一点一零伏；按这个顺序算，结果总是正的。
It tells you the polarity too: the more negative half-cell is the negative terminal.
它还告诉你极性： 较负的那个半电池是负极。
And there are two charge carriers: electrons in the wire, ions in the solution.
另外要注意有两种载流子：导线里是电子，溶液里是离子。
Line the standard electrode potentials up in order and you get the electrochemical series, with hydrogen at zero — a ranking of reactivity.
把所有标准电极电势按大小排好，就得到电化学序，氢按定义位于中间的零点——这也是反应活性的排序。
At the top are the most positive: fluorine, chlorine, bromine.
最上面是最正的：氟、氯、溴。
They are reduced very easily, so they are the strongest oxidising agents.
它们非常容易被还原，所以是最强的氧化剂。
At the bottom are the most negative: zinc, sodium — easily oxidised, the strongest reducing agents.
最下面是最负的：锌、钠——很容易被氧化，是最强的还原剂。
If the standard cell potential is positive, the reaction is feasible — that is the feasibility test.
如果标准电池电势是正的，反应就是可行的——这就是可行性判据。
Take bromine, plus one point zero seven, and the iron three, iron two pair, plus zero point seven seven.
以溴为例，正一点零七； 再看三价铁与二价铁这一对，正零点七七。
Bromine is more positive, so bromine is reduced and iron two is oxidised.
溴更正，所以溴被还原，二价铁被氧化。
Subtract: plus zero point three zero.
相减：正零点三零。
For the full equation, reverse the less positive half-equation, multiply so the electrons cancel, then add: bromine and two iron two ions give two bromide ions and two iron three ions.
要写出完整方程式：把较不正的半反应式倒过来写， 乘以适当倍数让电子抵消，然后相加：溴加上两个二价铁离子， 生成两个溴离子和两个三价铁离子。
Real solutions are rarely standard, so treat the half-cell as an equilibrium.
真实溶液很少处于标准状态，所以把半电池看成一个平衡。
Add more of the oxidised species and the potential becomes more positive. Add more of the reduced species and it becomes less positive.
增加氧化态物种，电势变得更正；增加还原态物种，电势就变得不那么正。
The Nernst equation puts a number on that: from the standard value, add zero point zero five nine over the number of electrons, times the logarithm of oxidised over reduced.
能斯特方程把这件事量化：从标准值出发，加上零点零五九除以电子数， 再乘以氧化态比还原态的对数。
Two cautions: oxidised on top, and count the electrons in that half-equation.
有两点要小心：氧化态写在上面； 电子数要数这一个半反应式里的。
One more.
再来一道真题。
A cobalt half-cell, standard value minus zero point two eight volts, but the cobalt ions are only zero point zero two moles per cubic decimetre.
一个钴半电池，标准值是负零点二八伏， 但钴离子浓度只有零点零二摩尔每立方分米。
Find the electrode potential, then the cell potential against standard dichromate.
求它的电极电势， 再求它与标准重铬酸盐电极组成电池的电池电势。
Two electrons are transferred, so the factor is zero point zero five nine over two.
转移的电子数是二， 所以系数是零点零五九除以二。
The logarithm of zero point zero two is minus one point seven, a correction of minus zero point zero five volts.
零点零二的对数是负一点七， 修正量是负零点零五伏。
The electrode potential is minus zero point three three volts.
于是电极电势是负零点三三伏。
Now the cell: more positive minus less positive gives one point six six volts.
再算电池：较正的减去较不正的，得到一点六六伏。
That beats the standard one point six one — diluting the cobalt made the cell stronger.
这比标准状态下的一点六一还要大——把钴稀释，反而让电池更强了。
Finally, a bridge back to energetics.
最后，架一座回到能量学的桥。
Free energy equals minus the moles of electrons, times the cell potential, times the Faraday constant.
自由能等于负的电子物质的量，乘以电池电势， 再乘以法拉第常量。
A past-paper case: plus one point six seven volts, two electrons transferred.
看一道真题：电池电势是正一点六七伏，转移两个电子。
Multiply out: minus three hundred and twenty two kilojoules per mole.
乘出来：负三百二十二千焦每摩尔。
Positive cell potential, negative free energy, feasible reaction — the two tests always agree.
电池电势为正、自由能为负、反应可行—— 这两种判据永远一致。
Three marks students throw away.
三个学生常丢的分。
First, the cell potential is always more positive minus less positive; never flip the sign of a half-cell value.
第一，电池电势永远是较正的减去较不正的； 不要去改半电池数值的正负号。
Second, if a question says define, give all three standard conditions: 298 K, one mole per cubic decimetre, and 100 kPa.
第二，如果题目要求下定义，三个标准条件都要写全： 298 K、一摩尔每立方分米、100 kPa。
Third, in the Nernst equation the oxidised species goes on top.
第三，能斯特方程里氧化态写在上面。
Get those right, and this topic is yours.
做对这三点，这个专题就是你的了。