Question 17.6

A Zn/Zn²⁺ concentration cell is constructed in which both electrodes are pure zinc. The Zn²⁺ concentration for one cell half is 1.0 M, for the other, 10^-2M. Is a voltage generated between the two cell halves? If so, how big is it and which electrode will be oxidized? If, on the other hand, you don't think any voltage will be produced, explain why not.

Answer 17.6

Yes, a voltage will be generated, and it can be calculated from the Nernst Equation, Eq. 17.19, or, assuming this is all happening at room temperature, Equation 17.20:

Delta V = (V₂⁰ - V₁⁰)-(0.0592/n)log(M₁ⁿ⁺/M₂ⁿ⁺)

The first term on the right-hand side goes to zero, while the second term evaluates to -0.0592/2 * log(0.01/1) = 0.0592 V. (We get the same answer whether we use Eq. 17.19 or 17.20.)

The zinc in the 0.01M solution will get oxidised, and its electrons will flow to the other half of the cell.

Question 17.14

A thick steel sheet of area 64,500 mm² is exposed to air near the ocean. After a one-year period, it was found to experience a weight loss of 485 gm due to corrosion. To what rate of corrosion, in mm/year, does this correspond ?

Answer 17.14

This could be solved using Equation 17.23. Personally, I don't like Equation 17.23 -- it introduces an unnecessary numerical factor K to permit you to use an inconsistent set of units, when the problem could be solved more straightforwardly by using a consistent set of units throughout.

We need to know the density of steel, which is about 7.8 g/cm³. So 485 gm of steel corresponds to 485 / 7.8 = 62.17 cm³. So, assuming that the steel is eaten away to the same depth over its entire surface, the depth it's eaten away to over the year is:

62.17 * 1000 / 64,500 mm = 0.963 mm/year.

(Your answer may be a bit higher or lower than this, depending on what value you take for the density of steel; anything in the range 0.94-0.97 is OK.)

Question 17.35

According to Table 17.3 (6th edition), the oxide coating that forms on silver should be nonprotective, but in fact silver does not oxidise appreciably in air at room temperature. How do you explain this apparent discrepancy?

Answer 17.35

According to Table 17.3, the PB ratio for silver oxide is 1.59, which is actually quite favourable -- exactly the same as beryllium, which is listed as protective, and better than lead, copper or iron. Still, let's accept that the oxide film is non-protective.

Silver does oxidise in air -- we refer to this as `tarnishing' -- and if you polish a piece of silver, you find that it grows dark again in a few weeks. So the explanation cannot be that the oxidation reaction is thermodynamically impossible.

As we see from Table 17.2, silver is very unreactive. So the oxidation reaction, although it occurs, does so very slowly.

Question 17.37

The following table gives weight-gain vs. time data for the oxidation of a certain metal at an elevated temperature:

W (mg/cm2)	Time (min)
6.16	100
8.59	250
12.72	1000

1. Determine whether the oxidation kinetics obey a linear, parabolic or logarithmic rate expression.
2. Compute W after a time of 5,000 minutes.

Answer 17.37

It's easy to show that these figures don't fit a linear relationship such as Eq. 17.32. We can also show that they don't fit a parabolic relationship such as Eq. 17.31. (Equation 17.31 is strange: obviously W=0 when t=0, so K₂ should always be zero.) So it seems like the expression must be logarithmic, and we just have to solve three equations of the form of Equation 17.33:

W = K₄ log(K₅t+K₆)

to find the three constants K.

Unfortunately, this is very tricky, despite having 3 equations and 3 unknowns. I've spent several hours working at an analytic solution, with no success. Maple will generate a solution, with careful handling: K4=7.305, K5=0.0535, K6 = 1.622. (Note that this doesn't give W=0 at t=0.)

Once we've got this solution, we substitute it back into Eq. 17.33 with t = 5000, to get W(5000) = 17.75 mg/cm³.

Question 18.11

Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is 1000 V/m. Under these circumstances how long does it take an electron to traverse a 25-mm length of crystal?

Answer 18.11

We obtain the drift velocity from Eq. 18.7:

v_d= mu_e E

The electron mobility in germanium may be found from Table 18.2; it's 0.38 m²/V-s. So the drift velocity is 0.38 * 1000 m/s = 380 m/s.

At this velocity, it would take an electron 0.025/380 s = 65.8 microseconds to cover a distance of 25 mm.

Question 18.14

1. Calculate the number of free electrons per cubic metre for silver, assuming there are 1.3 free electrons per silver atom. The electrical conductivity and density of silver are 6.8 * 10⁷ per ohm-metre and 10.5 gm/cm³ respectively.

2. Now calculate the electron mobility for silver.

Answer 18.14

We could start by finding the number of silver atoms in a cubic meter of silve:

A cubic meter of silver has a mass of 10,500 kg.

A kg-mole of silver weighs 107.87 kg.

So a cubic meter contains 10,500/107.87 kg-moles of silver.

A kg-mole of anything contains 6.023 * 10²⁶ atoms (or molecules).

So a cubic meter of silver contains 10500 * 6.023 * 10²⁶/107.87 atoms

So it contains 1.3 * 10500 * 6.023 * 10²⁶/107.87 = 7.62 * 10²⁸ electrons

Now we use Equation 18.8 to find the mobility:

mobility = sigma/ne = 6.8 * 10⁷/(7.62 * 10²⁸*1.6*10^-19)

= 0.00557 m²/V.s

Question 18.20

A cylindrical wire 2 mm in diameter is required to carry a current of 10 A with a maximum 0.01 V drop per 100 mm of wire. Which of the metals and alloys listed in Table 18.1 (6th edition) are candidates?

NOTE: In Callister, this question asks for a minimum 0.01 V drop per 100 mm, but this doesn't seem like a realistic requirement; if any voltage drop greater than 0.01 V per 100 mm is acceptable, you might as well leave the wire out entirely and increase the voltage until you get an arc.

Answer 18.20

We start from Equation 18.3:

rho = VA/Il

So if 0.01V/100 mm is the maximum permissible voltage drop, then rho must be less than

0.01 * pi * 0.001²/(0.1 * 10) = pi * 10^-8 ohms.metre

So the conductivity must be greater than (1/pi) * 10⁸ = 0.3183 * 10⁸ /ohm.m

So the top four metals in Table 18.1 -- silver, copper, gold and aluminium -- are candidates.

Question 18.53

At temperatures between 540 C and 727 C, the activation energy and pre-exponential for the diffusion coefficient of Na⁺ in NaCl are 173,000 J/mol and 4.0 * 10^-4 m²/s, respectively. Compute the mobility for an Na⁺ ion at 600 C.

Answer 18.53

This takes us back to Equation 5.8:

D = D₀exp(-Q_d/(RT))

So D = 4.0 * 10^-4 * exp(-173,000/(8.31*873))=1.76 * 10^-14 m²/sec

Now the mobility is given by Equation 18.20:

mobility = neD/(kT) = 1.602 *10^-19 * 1.76 * 10^-14/(1.38 * 10^-23* 873)

= 2.34 * 10^-13 m²/sec