1973-04-07 — Page 18

真二第張五第二日五初月三年壯発展夏

WAH KIU YAT PO

報日橋單

六期星

5. Faraday's laws of electrolysis

First law: the mass of any

1973英文中學會考試題預習專欄

BY AN ★ AREA

物理科

Physics (24)

The potentiometer

Combining (2) & (3)

(廿四)

or r

(~ - 1)R.

A potentiometer consists of a uniform wire about a metre long maintained at a constant potential

gradient by an accumulator.

accumulator

METRE WIRE

Since the wire is unifom, the p.d. between the points A & C, V is proportional to the length

AC

AC

If AC = l, VAC & l.

Application of potentiometer

potent Comparison of B.]

To compare the emfs of 2 cells, the circuit in fig 2. employed.

HE

In the circuit shown, G is a galvanometer and P is the protect- ive resistance. It is required td compare the emfs of the cells of

•&

First close the key K and obtained the balance point for the cell E1. Suppose the point is at C. This means that the potential at C is the same as that at F So the length AC of the Wire exactly balances the emf E. Use the fundamental equation (1),

ке

where Xy

length of AC. Then K is opened and K2 closed. Repeat the process until the galvanometer gives no deflection. i.e. the balance point is obtained Suppose this time is at D. Then

the potential at D is the same as the potential at F.

Again E2- Kl2

By division

Hence the ratio of emfs of the jcells can be determined.

(B) Measurement of emf and inter

1 resistance of a cell.

To find the emf and the internal resistance of an unknown call X a standard cell whose emf is known is used and the circuit employed is in fig 3.

In the circuit, R 15 a known resistance. Suppose C is the balance point when the unknown cell X is inserted with R not in position.

Then E KL

- AC.

With R in position, the balance. point will be different, say at D. Then V = K}'

£ = AD.

Then

On the other hand,

E = 1(R + r)

r internal resistance of the

Moell.

But the current i in the is given by Ohm's law as

Hence = R+

V

circuit

i =

(3)

Hence the internal resistance determined.

Next will the standard cell in place of the unknown cell X, a new balance point is obtained, Say at F.

Then

Thus

lo

lo = AF.

e.g. 1. The emf of a cell is balanced by the fall in potential along 150 cm. of a potentiometer wire. When the cell is shunted by a resistance of 14 ohms the

What length required is 140 cm.

is the internal resistance of the cell? Let r be the internal resistance of the cell. Apply the formula in B.

150% 110

Ohm.

1)X 14

3. Bridge circuit.

A common method of determin- ing the unknown resistance is by means of a Wheatstone bridge, sometimes called a metre bridge. The bridge consists of a wire 1 metre long. The unknown resistan- ce X and a known resistance R are connected, as shown in fig 4

When the slider is at the point C, the wire is divided into two parts of resistance R

and RCB

If C is the balance point, the potential difference between D& C is zero, 1.e. they are at the same potentials. An alternate. form of the circuit shown in fig 4 is given in fig 5.

Then VAC

and VBC

VAD

BD

AC 12X

11 RCB

BAC

By division,

BCB

Since the wire is uniform, the length is proportional to the resistance, i.e. Rc

Rop

Let AC =

Then X

length of AC length of BC.

OB

4. Electrolysis

The chemical changes which occur when a current passes. through a conducting solution called an electrolyte is called electrolysis. The electrolyte used is an acid, a base or a salt. solution. The plates or wires lip into the electrolyte to con- nect it to the circuit are called electrodes. The electrode connect ed to the positive terminal of a cell is called the anode, the other connected to the negative terminal of the cell is called the cathode. The whole arrange- ment is called a voltameter.

日七月四年三七九一膳公年二十六國民華中有教僑華

the south pole. The field of a

flat circular coil is quite differ

-ent from that of a solenoid and is given in fig 8

substance liberated in electroly-

sis is proportional to the

quantity of

liberated it.

Second Law Lectricity that

the masses of differ-

ent substances liberated in electrolysis by the same current are proportional to their chemical equivalents. Me

In mathematical form, the first law can be written as

m = zit

where m is the mass liberated in. electrolysis, i is the current in the circuit, t is the time in electrolysis and z is called the electrochemical equivalent of the element. • The unit of z is in gms per coulomb. The second law is

wt of element of element eq. wt.

The experimental verification of these 2 laws can be found in many text books and will not be repeat- ed.

e.g. 2. A copper voltameter consist -s of two parallel copper plates 6 cm apart and 1 m. square immersed in

solution of

Cuso

resistivity 1.2 ohm/cm. What must be the p.d. between the plates in order to deposit 480 gm. of copper on the cathode in one hour. (electrochemical equivalent. of copper is 0.00033 gm/coul.) Applying

m = zit

480 0.00033 x 3600

amp.

resistance of the cell,

where

106100

resistivity.

Hend

x 100 ohms.

iR the p.d. V

480 x 1.

00033 x 3600 x 100 x 100 0.3 V. (approx.

6. The Faraday

The quantity of electricity which will deposit one gram-equi- valent of any substance is called faraday. It is equal: to 96500 coul. It can be observed that the electrochemical equivalent of any substance is given by-

z = equivalent weight of substance

96500

7. Magnetic effect of current

Oersted discovered that a current, carrying conductor pro- duces a magnetic field round it. The magnetic field pattem in a plane of a long straight wire/ carrying current i is a series of concentric circles. The direction of the magnetic field is given by the Right-Hand-Grip rule which states that if the direction of the thumb of righthand is the current direction. Then the direction of fingers which grip the wire is the direction of field lines. Thus as shown in fig 6.

The magnetic field direction for a wire carrying current in the upward direction is indicated by the arrows. On the other hand, the field of a current-carrying solenord is similar to that of a bar magnet..

Thus a solenoid carrying current. i behaves as a bar magnet. The current which flows in the anti- clockwise direction when viewed from one end is the north pole while the current which flows in the clockwire direction when viewed from the other end gives

It was found experimentally that the magnetic field strength pro- duced by a current-carrying con- ductor is proportional to the strength of the current in the conductor,

8. Eleming's left hand rule.

Now consider a current in a wire which is placed in a uniform magnetic field as shown in fig 9

N

nevemy

The current is in the downward direction. It is observed that when the circuit is closed, the wire swings forward. If the current is reversed, the wire swings backward. The direction

wire is related of motion of

of current and the direction of the magnetic field by the famous Fleming's hand rule stated as: put out the thumb, the fore finger and the middle finger of the left hand so that they are at right angles to each other. Then the thumb

Ves the direction of motion,

forefinger gives the direction the magnetic field and the middle finger gives the direction. of current.

to the dire the w

gi

of

Exercise

motim

field

current

1. A 155 cm length of potentio- meter wire balances the emf of a cell on open circuit and a 135 cm. length balances the p.d. between the terminals of the same cell when it has a conductor of re- sistance 8 ohms connected between. its terminals. What is the internal resistance of the cell?

2. A circuit consists of a battery of emf 10 V. & of negligible, internal resistance and two corts in series each of resistance 500 ohms. A voltmeter of resistance 1000A is connected in parallel across one of the 500 corts. Find the reading of the voltmeter.

3. What fraction of the electri cal energy is converted into chemical energy when a 6 V. battery is

water?

1.04 x 10

to electrolyse

e. of hydrogen is

The combust- of hydrogen to form water liberates 147 KJ.)

ion of 1

4. Two copper voltameters are in series and one of them is joined in parallel with a silver voltameter. The rates of deposit- ion in the copper voltameters are in the ratio 1 to 3 and that 4 gm. of silver are deposited per hour. Find the total amount of copper deposited in an hour given that the e.c.e. of silver and copper are 108 and 31-8 respectively.