practical report

CIE 5100 Repair and maintenance of construction materials


Practical ‘Corrosion’

Petrica Banea 4125525 Coen Smits 4039688

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Question 1. Give a summary table of all measurement data and observations schematically of both the agar experiments, the oxygen cell and the i-E values of zinc and steel. Also note the observations of the passivation of the nail. Set-up Visual result

A Copper rod + steel nail in agar The agar turned blue at the nail and pink at the copper rod

B Zinc bar + steel nail in agar The agar turned pink at the nail and white at the zinc bar

C Platinum wire + steel nail in agar The agar turned blue at the nail and pink at the platinum wire

D Platinum wire + steel nail in agar, not short-circuited and 0,5V over them

The agar turned half blue / half pink at the nail and stayed yellow at the platinum wire

E Platinum wire + steel nail in agar, not short-circuited and 2V over them

The agar turned pink at the nail and green/blue at the platinum wire

F Hammered nail with the head down in agar

The agar turned blue around the hammered head and pink around the rest of the nail

G Passivated and filed nail in agar At the filed tip, the agar turned blue, at the unfiled and passivated part, the agar turned pink

H Two nails in sea water The water at both sides became a bit rusty.

Measurements of potential differences between 2 nails in salt water after adding pure oxygen to the water: T (s) Potential difference 0 -123mV 180 -52,5mV 300 -42,7mV 420 -36,1mV 540 -32,1mV

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Measurements of the potential difference in seawater between a zinc- and a reference electrode:

Potential (V)

Zn Zn(changed polarity) Amperage 0mA -1,065 -1,081 2mA -1,042 -1,117 4mA -1,039 -1,184 6mA -1,033 -1,203 8mA -1,029 -1,359 10mA -1,024 -1,394 15mA -1,016 -1,434 20mA -1,009 -1,436 25mA -1,005 -1,445

30mA -1,001

we couldn’t read this because we couldn’t reach the 30mA to do the reading

Same for the steel electrode and the reference material:

Curent (V)

Fe Fe(changed polarity) Amperage 0mA -0,671 -0,711 2mA -0,631 -0,846 4mA -0,613 -0,897 6mA -0,601 -0,918 8mA -0,59 -0,935 10mA -0,582 -0,96 15mA -0,566 -1,168 20mA -0,553 -1,22 25mA -0,543 -1,244 30mA -0,534 -1,264

We also passivated a nail by putting a cleaned nail in Nitric acid. When we did this, for a very short moment some yellow/brown gas appeared and then nothing happened anymore. This seemed to be the passivation process.

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Question 2. Answer all the questions from the “Experimental Procedure” in the right order Question: Under what conditions will they cause a color change? - Phenolphtalein will turn pink when the pH of the solution rises to between 8,2 and

12. - Potassium ferricyanide turns blue in the presence of Fe2+ ions Question: What will be roughly the temperature of the nail? 900 degrees Celsius Question: What gas will evolve during the etching of the nails? Give the reaction equation. Hydrogen. 2Fe (s) + 6HCl (aq) 2FeCl3 (aq) + 3H2 (g) Question: Which reaction will occur during passivation of the nail? Fe + 4HNO3 Fe(NO3)3 + 2H2O + NO Question: Which nail has the lowest potential after O2 gas is blown through the solution? Which reactions will occur during the O2 gas bubbling? Which nail is then the cathode? Negative values The nail at the red wire has the lowest potential and is therefore the anode. The nail at the black wire is the cathode. At the anode, oxidation will occur: Fe Fe2+ + 2e At the cathode, reduction happens: O2 + 2H2O + 4e 4OH- Question: Which form of corrosion prevention is taking place now? Cathodic protection by using a sacrificial metal (zinc) which corrodes easier than the main metal (steel).

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Question 3. Give a complete explanation of the observed corrosion phenomena with the “agar” experiments. Setup A In setup A a steel nail and a copper rod were short-circuited placed in the agar. The result after 2 days was pink colored agar around the copper rod and dark blue agar around the nail. The indicators used show that around the copper, reduction occurred (the amount of OH- increased and rose the pH) and around the nail oxidation occurred (Potassium ferricyanide proofed the existence of Fe2+ ions). At the anode (steel, oxidation): Fe Fe2+ + 2e At the cathode (copper, reduction): O2 + 2H2O +4e 4OH- Setup B In setup B, a zinc bar replaced the copper rod. Since zinc is less noble than steel, oxidation in this case occurred at the zinc bar and reduction occurred at the steel nail. The agar around the steel nailed turned pink, proofing the existence of OH- and turned white around the zinc bar. The white color is caused by Zn2+ ions, which were lost to the solution. At the anode (zinc, oxidation): Zn Zn2+ + 2e At the cathode (steel, reduction): O2 + 2H2O +4e 4OH- Setup C In setup C, the zinc bar was replaced. This time a platinum wire was short-circuited with the nail and both were put in the agar. Reduction occurred around the platinum wire, oxidation occurred at the steel nail. At the anode (steel, oxidation): Fe Fe2+ + 2e At the cathode (platinum, reduction): O2 + 2H2O +4e 4OH-

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Setup D For this setup, we used the same setup as in setup C, but now removed the short circuit plug and a voltage of 0,5V was connected to the 2 electrodes. Because we chose the nail to be the cathode (black pole), we expected a cathodic reaction at this side. But since the platinum wire is more noble, this side could also be expected to be the cathodic side. After 2 days, a bit of both occurred. At the surface of the agar reduction- and at the tip of the nail, oxidation occurred. The platinum wire is no longer protected now, since no OH- appeared in the solution. At the steel nail, both reduction and oxidation: Fe Fe2+ + 2e O2 + 2H2O +4e 4OH- At the platinum wire (clearly the cathode): Pt Pt2+ + 2e Setup E Same as setup D, now with a higher voltage (2V instead of 0,5V). This time, the steel nail is fully protected, since no only the pink colored indicator appeared. At the cathode (steel, reduction): O2 + 2H2O +4e 4OH- At the anode (platinum, oxidation) Pt Pt2+ + 2e Setup F In this setup, only a nail is placed in the agar. The nail is hammered at the head, and placed upside down into the agar. In this case, the nail is it’s own anode and cathode at once. As seen in the picture on the right, the killed steel oxidated so strong that the unkilled steel is protected by it. At the tip (anodic, oxidation): Fe Fe2+ + 2e At the agar surface (cathodic, reduction): O2 + 2H2O +4e 4OH-

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Setup G Same as setup H, but now we placed a passivated nail with a filed tip. By passivation, a protective layer appeared around the nail, with filing we removed this at the tip of the nail. The result is the same as in setup H. At the tip (anodic, oxidation): Fe Fe2+ + 2e At the agar surface (cathodic, reduction): O2 + 2H2O +4e 4OH-

Setup H In this case we placed 2 nails in seawater and short-circuited them. Since there were no indicators in the seawater, oxidation or reduction can’t be proofed, but it was clear that oxidation at both nails occurred since rust was in the water at both nails.

1. Fe Fe2+ + 2e 2. O2 + 2H2O + 4e 4OH- 3. Fe2+ + 2OH- Fe(OH)2 4. 4 Fe(OH)2 + O2 2 (Fe2O3 x H2O) + 2H2O

Fe2O3 is rust, which we saw. Question 4. Give an explanation of the oxygen-cell by answering the next questions.

4a. Which reactions take place before O2 gas is blown through? These reactions are as described in the previous question:


4b. Which reactions take place during the O2 supply? The reduction process at the cathodic nail accelerates (eq. 2 at setup H), creating more free electrons. These electrons accelerate the oxidation process at the other (anodic) nail. 4c. Which nail has directly after the O2 supply the highest potential and which one is the anode?

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The nail next to the O2 supply is the cathode and has the highest potential. The other nail is the anode. 4d. Explain the change of potential after the start of blowing with oxygen gas. The free oxygen in at the cathode decreases, slowing down the creation of free electrons and the system is trying to create an equilibrium. Question 5. The measured data for the electrode potential of zinc and steel in seawater can be read in the following tables:

Zn (zinc)

Current value i[mA]

Natural Logarithm

of i lni

Measured Electrode Potential [V] Normal polarity Inversed Polarity

0 - -1.065 -1.081 3.2 1.163 -1.042 -1.117 4 1.386 -1.039 -1.184 6 1.792 -1.033 -1.203 8 2.079 -1.029 -1.359

10 2.303 -1.024 -1.394 15 2.708 -1.016 -1.434 20 2.996 -1.009 -1.436 25 3.219 -1.005 -1.445 30 3.401 -1.001 -1.449

Table 01

Fe (steel)

Current value i[mA]

Natural Logarithm

of i lni

Measured Electrode Potential [V] Normal polarity Inversed Polarity

0 - -0.671 -0.711 2 0.693 -0.631 -0.846 4 1.386 -0.613 -0.897 6 1.792 -0.601 -0.918 8 2.079 -0.590 -0.935

10 2.303 -0.582 -0.960 15 2.708 -0.566 -1.168 20 2.996 -0.553 -1.220 25 3.219 -0.543 -1.244 30 3.401 -0.534 -1.264

Table 02

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Question 6. a) According to the Table 01 and Table 02, that contains the measured data for the Zn (zinc) electrode and Fe (steel) electrode, respectively, the Evans diagrams are constructed as they are shown in the graphs below.

Figure 01. Evans diagram for Zn electrode in seawater

-1.600

-1.400

-1.200

-1.000

-0.800

-0.600

-0.400

-0.200

0.0000.000 1.000 2.000 3.000 4.000 5.000

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corrosion current (lni) [mA] →

Zn-electrode: scatter data

Zn(A)

Zn(K)

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Figure 02. Evans diagram for Fe electrode in seawater b) In order to find the mixed (corrosion) potential for Zn and Fe electrode from the diagrams, we have to linearly interpolate the existing lines. The following graphs show the intersections of the cathodic and anodic lines through which we found the exact values of the mixed potential as they are shown in figure 03 & 04 and later we compared them with the measured values. For this purpose we used the Excel graphs tool in order to plot the interpolated line of the scattered measured data. This can be achieved by selecting the option ‘Add Trendline’ to the displayed scatters. This command is translated into a line, that has the linear equation of the ‘y = ax + b’ form. The coordinates of the intersection point of the two Trendlines were calculated with Excel and they represent the mixed potential (y-value – Emix) and the corrosion current (x-value – icorr).

-1.400

-1.200

-1.000

-0.800

-0.600

-0.400

-0.200

0.0000.000 1.000 2.000 3.000 4.000 5.000

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Fe-electrode: scatter data

Fe(A)

Fe(K)

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Figure 03. Interpolation of the cathodic and anodic line for Zn-electrode From the graph we can extract the linear equation of the anodic and cathodic lines and with the help of the ‘Goal Seek…’ function in Data/What-if Analysis/, we are able to calculate the coordinates of the intersection point between the two lines. This intersection of the anodic and cathodic line for each metal will supply the mix potential [Emix] and the corrosion current [icorr]. Therefore the values are:

- the equation of the anodic line: y1 = 0.0186x - 1.0656

- the equation of the cathodic line: y2 = -0.153x - 0.9778

x-value y1-value y2-value y1-y2

0.512 -1.056083217 -1.056083217 0

y-value Emix = -1.056083217 V x-value ln(icorr) = 0.512 icorr = eln(icorr) = 1.668049553 mA

y = 0.0186x - 1.0656

y = -0.153x - 0.9778

-1.800

-1.600

-1.400

-1.200

-1.000

-0.800

-0.600

-0.400

-0.200

0.000-3.000 -2.000 -1.000 0.000 1.000 2.000 3.000 4.000 5.000

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Zn-electrode: interpolated data

Zn(A)

Zn(K)

∩ (0.512 , -1.05608)

Emix = -1.05608

ln(icorr) = 0.512

Linear (Zn(A))

Linear (Zn(K))

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Figure 04. Interpolation of the cathodic and anodic line for Fe-electrode For the Fe-electrode the procedure is similar to the one for Zn-electrode. Therefore the Emix and icorr are as follows:

- the equation of the anodic line: y1 = 0.0364x - 0.6625

- the equation of the cathodic line: y2 = -0.1773x - 0.6448


0.083 -0.659485119 -0.659485119 0

y-value Emix = -0.659485119 V

x-value ln(icorr) = 0.083 icorr = eln(icorr) = 1.086353193 mA Calculated corrosion potential values EmixZn = -1.0561 V icorrZn = 1.6680 mA

EmixFe = -0.6595 V icorrFe = 1.0864 mA

Measured corrosion potential values EmixZn = -1.0510 V

EmixFe = -0.7030 V

Comparing the results between the mixed corrosion potential and the measured values for each metal, we can notice a slightly difference between them.

y = 0.0364x - 0.6625

y = -0.1773x - 0.6448

-1.600

-1.400

-1.200

-1.000

-0.800

-0.600

-0.400

-0.200

0.000-4.000 -2.000 0.000 2.000 4.000 6.000

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Fe-electrode: interpolated data

Fe(A)

Fe(K)

∩ (0.083 , -0.65949)

Emix = -0.65949

ln(icorr) = 0.083

Linear (Fe(A))

Linear (Fe(K))

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Question 7. The corrosion rate (in mm/year) of the zinc (Zn) electrode is determined as follow: For the mix potential of Zn EmixZn = - 1.0561 V, the corrosion current is icorrZn = 1.6680 mA With the Faraday Law we can calculate the corrosion rate for Zn. z – The valence of the metal ion M – Molar mass ρZn – Zn density F – Faraday’s constant SZn – the surface of the Zn-electrode W – The amount of deposited or dissolved metal t – Time 𝑧𝑍𝑛 = 2; 𝑀𝑍𝑛 = 65.409

𝑔𝑚𝑜𝑙

ρ𝑍𝑛 = 7.14𝑔𝑐𝑚3 ; 𝐹 = 96484.56

𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑚𝑜𝑙

𝑆𝑍𝑛 = 12.8𝑐𝑚2; 𝑡 = 60 × 60 × 24 × 365 = 31536000𝑠

𝑊𝑍𝑛 =𝑀𝑧 × 𝑖𝑐𝑜𝑟𝑟 × 𝑡

𝐹=

65.4092 × 0.001668 × 31536000

96484.56= 17.83

𝑔𝑦𝑒𝑎𝑟

𝑣𝑍𝑛 =𝑊𝑍𝑛

ρ𝑍𝑛=

17.837.14

= 2.499𝑐𝑚3

𝑦𝑒𝑎𝑟

The corrosion speed is:

𝑠𝑍𝑛 =𝑣𝑍𝑛𝑆𝑍𝑛

=2.49912.8

× 10 = 𝟏.𝟗𝟓𝒎𝒎𝒚𝒆𝒂𝒓

The corrosion rate (in mm/year) of the steel (Fe) electrode is determined as follow: For the mix potential of Fe EmixZn = - 0.6595 V, the corrosion current is icorrZn = 1.0864 mA With the Faraday Law we can calculate the corrosion rate for Fe. z – The valence of the metal ion M – Molar mass ρFe – Fe density F – Faraday’s constant SFe – the surface of the Fe-electrode W – The amount of deposited or dissolved metal t – Time 𝑧𝐹𝑒 = 2; 𝑀𝐹𝑒 = 55.845

𝑔𝑚𝑜𝑙

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ρ𝐹𝑒 = 7.86𝑔𝑐𝑚3 ; 𝐹 = 96484.56


𝑆𝐹𝑒 = 12.8𝑐𝑚2; 𝑡 = 60 × 60 × 24 × 365 = 31536000𝑠

𝑊𝐹𝑒 =𝑀𝑧 × 𝑖𝑐𝑜𝑟𝑟 × 𝑡

𝐹=

55.845 2 × 0.0010864 × 31536000

96484.56= 9.91


𝑣𝐹𝑒 =𝑊𝐹𝑒

ρ𝐹𝑒=

9.917.86

= 1.261 𝑐𝑚3

𝑦𝑒𝑎𝑟


𝑠𝐹𝑒 =𝑣𝐹𝑒𝑆𝐹𝑒

=1.26112.8

× 10 = 𝟎.𝟗𝟖𝟓𝒎𝒎𝒚𝒆𝒂𝒓

Question 8. The corrosion rate (in mm/year) of the steel (Fe) electrode when it is short-circuited with the zinc (Zn) electrode is as follows: In this case, the least noble metal is then the anode. Therefore, the zinc (Zn) becomes the anodic and steel (Fe) becomes the cathodic. Intersection can be read in the figure 05.

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Figure 05. Interpolation of the cathodic and anodic line for Fe-electrode Using the same procedure as for the Question 6, we then can find out the intersection points of the anodic(Zn) and cathodic(Fe) line.

- the equation of the anodic line, Zn(A): y1 = 0.0186x - 1.0656

- the equation of the cathodic line, Fe(K): y2 = -0.1773x - 0.6448


y = 0.0186x - 1.0656

y = -0.153x - 0.9778

y = 0.0364x - 0.6625

y = -0.1773x - 0.6448

-2.000

-1.500

-1.000

-0.500

0.000

0.500

1.000

1.500-12.000 -10.000 -8.000 -6.000 -4.000 -2.000 0.000 2.000 4.000 6.000

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Zn(A)-Fe(K)-Zn(K)-Fe(A)

Zn(A) Zn(K)

Fe(A) Fe(K)

ln(icorr)Zn(A)Fe(K) = 2.148 EmixZn(A)Fe(K) = -1.0256

ln(icorr)Fe(A) = -9.977 ∩Zn(A)Fe(K) (2.148 , -1.0256)

∩Fe(A) (-9.977 , -1.0256) Linear (Zn(A))

Linear (Zn(K)) Linear (Fe(A))

Linear (Fe(K))

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2.148 -1.0256 -1.0256 0

y-value Emix = -1.0256 V x-value ln(icorr) = 2.148 icorr = eln(icorr) = 8.5680 mA

From this intersection point, we can then deduce the corrosion current on the anodic(Fe) line.

- the equation of the anodic line, Fe(A): y = 0.0364x -0.6625

From this equation we can find the x-value corresponding to the corrosion current of the Fe(A), therefore x = (y+0.6625)/0.0364 = -9.977

Emix = -1.0256 V

x-value y-value

-9.977 -1.0256

y-value Emix = -1.0256 V x-value ln(icorr) = -9.977 icorr = eln(icorr) = 0.000046477 mA

With the Faraday Law we can calculate the corrosion rate for Fe. z – The valence of the metal ion M – Molar mass ρFe – Fe density F – Faraday’s constant SFe – the surface of the Fe-electrode W – The amount of deposited or dissolved metal t – Time 𝑧𝐹𝑒 = 2; 𝑀𝐹𝑒 = 55.845

𝑔𝑚𝑜𝑙

ρ𝐹𝑒 = 7.86𝑔𝑐𝑚3 ; 𝐹 = 96484.56


𝑆𝐹𝑒 = 12.8𝑐𝑚2; 𝑡 = 60 × 60 × 24 × 365 = 31536000𝑠


𝐹=

55.845 2 × 0.000000046477 × 31536000

96484.56= 0.000424



ρ𝐹𝑒=

0.0004247.86

= 0.0000539 𝑐𝑚3

𝑦𝑒𝑎𝑟



=0.0000539

12.8× 10 = 𝟎.𝟎𝟎𝟎𝟎𝟒𝟐𝟐

𝒎𝒎𝒚𝒆𝒂𝒓

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Conclusion: After the connection of Fe with Zn, the corrosion rate of steel decreases with a factor of 23348 times (0.985/0.0000422). The method is proving very effectively for corrosion prevention and it shows the importance of Zn as a sacrificial metal for the longer lasting life of steel in corrosive medium. Question 9.

- In order to calculate after how many days the zinc (Zn) electrode has been sacrificed we use the Faraday’s law using t as the unknown value.

For the mix potential of Zn EmixZn = - 1.0256 V, the corrosion current is icorrZn = 8.568 mA With the Faraday Law we can calculate the corrosion rate for Zn. z – The valence of the metal ion M – Molar mass ρZn – Zn density F – Faraday’s constant SZn – the surface of the Zn-electrode W – The amount of deposited or dissolved metal t – Time 𝑧𝑍𝑛 = 2; 𝑀𝑍𝑛 = 65.409

𝑔𝑚𝑜𝑙

ρ𝑍𝑛 = 7.14𝑔𝑐𝑚3 ; 𝐹 = 96484.56


𝑆𝑍𝑛 = 12.8𝑐𝑚2; 𝑡 = 60 × 60 × 24 = 86400𝑠

𝑊𝑍𝑛 =𝑀𝑧 × 𝑖𝑐𝑜𝑟𝑟 × 𝑡

𝐹=

65.4092 × 8.568 × 10−3 × 86400

96484.56= 0.251

𝑔𝑑𝑎𝑦

𝑣𝑍𝑛 =𝑊𝑍𝑛

ρ𝑍𝑛=

0.2517.14

= 0.0351𝑐𝑚3

𝑑𝑎𝑦


𝑠𝑍𝑛 =𝑣𝑍𝑛𝑆𝑍𝑛

=2.49912.8

× 10 = 0.0274𝒎𝒎𝒅𝒂𝒚

Considering that per day the Zn-electrode is corroding by 0.0351 cm3 and the total volume of the electrode is 2 cm3, this will take 56.98 days for the Zn-electrode to completely dissolve. Similarly for the Fe-electrode using the results determined in question 8, (assuming that the surface remains constant during corrosion), with a dissolve ratio of𝟎.𝟎𝟎𝟎𝟎𝟓𝟑𝟗 𝐜𝐦

𝟑

𝐲𝐞𝐚𝐫,

the Fe-electrode, which has the same dimensions as the Zn one and will dissolve in approximately 37105 years.

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The results presented above are an example of galvanic corrosion in which one metal (in our case Zn) is sacrificed to protect another metal (Fe) and it can be notice the difference between the results from question 7 and question 9, when due to the short circuiting the Fe and Zn, the life span of Fe-electrode is increasing considerably. Question 10. a) Reducing the anodic surface by a factor of 3 means that the current ‘i’ will get a value of ‘i/3’. This means that the anodic line will get shifted with ln(3). ln �1

3𝑖� = ln �1

3� + ln(𝑖) = ln(𝑖) − 1.1

This can be noticed in the following new Evans diagram. EmixFe = - 0.617 V, ln(icorrFe) = -0.098 → icorrFe = e-0.098= 0.9066 mA b) If we reduce the cathodic surface with the same coefficient of ln(3) we will see in the below Evans diagram, the shifted cathodic line, from which we extrapolate the new mixed potential and corrosive current. EmixFe = - 0.645 V, ln(icorrFe) = -0.809 → icorrFe = e-0.809= 0.4453 mA c) The corrosion of the steel electrode is a cathodic-controlled process. This is obvious because of the high steepness of the cathode compared to the anode.

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Figure 06. Evans diagram of the shifted cathodic and anodic line for Fe-electrode Question 11. Calculate the molar volume of Fe2O3 and iron. Is the volume change such that after oxidation of the iron a good adherence between the metal and the oxide will occur? Vm = M / ρ Iron has an atomic weight of 55,8352 𝑔

𝑚𝑜𝑙

Oxygen has an atomic weight of 15,99943 𝑔𝑚𝑜𝑙

2 iron atoms and 3 oxygen atoms give molar mass M for Fe2Oe of 159,6887 𝑔

𝑚𝑜𝑙.

The mass density ρ of Iron is 7,874 𝑔

𝑐𝑚3 The mass density ρ of Oxygen is 0,001429 𝑔

𝑐𝑚3 The molar volume of Fe2O3 = 30,4 𝑐𝑚

3

𝑚𝑜𝑙

The molar volume of iron = 7,092 𝑐𝑚3

𝑚𝑜𝑙

y = 0.0364x - 0.6225

-1.600

-1.400

-1.200

-1.000

-0.800

-0.600

-0.400

-0.200

0.000-4.000 -2.000 0.000 2.000 4.000

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Fe-electrode: Shifted anodic and cathodic lines with ln(3)

Fe(A)-ln(3)

Fe(A)

Fe(K)

Fe(K)-ln(3)

Linear (Fe(A)-ln(3))

Linear (Fe(A))

Linear (Fe(K))

Linear (Fe(K)-ln(3))

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𝑀𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝐹𝑒2𝑂3𝑀𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝐹𝑒

=30,4

7,092= 4,29

4,29 > 1,8 Too large, so the oxide layer easily spalls from the surface. Result: poor protection. Question 12. Also calculate the molar volume of copper and CuO. What is your conclusion about the protective capability of the oxide? Vm = M / ρ Copper has an atomic weight of 63,546 𝑔

𝑚𝑜𝑙

Oxygen has an atomic weight of 15,99943 𝑔𝑚𝑜𝑙

CuO = 63,546 + 15,99 = 79,545 𝑔

𝑚𝑜𝑙

Mass density of CuO: ρ = 6,31 𝑔𝑐𝑚3

Molar volume of CuO = 12,606 𝑐𝑚3

𝑚𝑜𝑙

Mass density of copper ρ = 8,95 𝑔

𝑐𝑚3

The molar volume of copper = 7,092 𝑐𝑚3

𝑚𝑜𝑙

𝑀𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝐶𝑢𝑂𝑀𝑜𝑙𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝐶𝑢

=12,6067,092

= 1,78

1,2 < 1,78 < 1,8 This means the oxide layer is compact and continuous and it gives a good protection. Question 13. What are the reaction equations for the corrosion of steel in surface water and the corrosion of zinc in diluted hydrochloric acid? For the steel, the same happens as explained before:


For the zinc, the reaction formula describes the oxidation of zinc:

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Zn (s) + 2HCl (aq) ZnCL2 (aq) + H2 (g) Question 14. a) Mix potential Emix = - 0.45 V Corrosion current icorr = 10-4 A b) 75% of anodic regions covered with inhibitor it means that the current i is ‘1/4i’, so log(i) is log(i/4). y = a (logx) + b (Emix = a log(i) + b) Anode: -0.45 = a log(10-4/4) + b a = 0.1 -0.55 = a log(10-5/4) + b b = 0.0102 Cathode: -0.45 = a log(10-4) + b a = -0.1 -0.55 = a log(10-3) + b b = -0.85 Anode: Emix = 0.1 log(i) + 0.0102 Emix = -0.42 V Cathode: Emix = -0.1 log(i) -0.85 icorr = 5×10-5A c) Similar to the question above, but for this time the anode is unchanged, the function of the anode is: Anode: -0.45 = a log(10-4) + b a = 0.1 -0.55 = a log(10-5) + b b = 0.05 Emix = 0.1 log(i) + 0.05 Emix = -0.6V -0.6 = 0.1× log(i) – 0.05 → log(icorr) = -5.5 → icorr = 10-5.5 = 3.162×10-6 A 𝑧𝐹𝑒 = 2; 𝑀𝐹𝑒 = 55.845

𝑔𝑚𝑜𝑙

ρ𝐹𝑒 = 7.86𝑔𝑐𝑚3 ; 𝐹 = 96484.56


𝑆𝐹𝑒 = 4 × 5 × 4 + 4 × 4 × 2 = 112𝑐𝑚2; 𝑡 = 60 × 60 × 24 × 365 = 31536000𝑠


𝐹=

55.845 2 × 3.162 × 10−6 × 31536000

96484.56= 0.0289


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ρ𝐹𝑒=

0.02897.86

= 0.0037 𝑐𝑚3

𝑦𝑒𝑎𝑟



=0.0037

112× 10 = 𝟎.𝟎𝟎𝟎𝟑𝟑


d) The electric current that the power source should deliver, such that the mixed potential should be maintained at -0.60V is: 𝑖𝑛𝑒𝑡𝐹𝑒 = │𝑖𝑎𝑛𝑜𝑑𝑒𝐹𝑒 − 𝑖𝑐𝑎𝑡ℎ𝑜𝑑𝑒𝐹𝑒 │ Using the Evan diagram from figure xx, we then determine the values of ianodeFe and icathodeFe

Figure 07. Evans diagram of the cathodic and anodic line of Fe-electrode Fe(A) at -0.60V yields: ianodeFe =10-5.5A Fe(K) at -0.60V yields: icathodeFe =10-2.5A 𝑖𝑛𝑒𝑡𝐹𝑒 = │𝑖𝑎𝑛𝑜𝑑𝑒𝐹𝑒 − 𝑖𝑐𝑎𝑡ℎ𝑜𝑑𝑒𝐹𝑒 │ = │10−5.5 − 10−2.5│ = 0.00316A Because the anodic current is negligible compared to the cathodic one, to keep the steel plate at -0.6V a cathodic current of 0.00316A should be flowing.

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e) If the Fe plate is short circuited with the Al one, then the data collected from the Evans diagram are as following: Mix potential Emix = - 0.65 V Corrosion current icorr = 10-2 A But, in the case of the short circuit, the anodic current of the Fe-plate will have the corresponding value of icorr = 10-6 A. Using Faraday’s Law, we then calculate the corrosive rate of Fe plate. 𝑧𝐹𝑒 = 2; 𝑀𝐹𝑒 = 55.845

𝑔𝑚𝑜𝑙

ρ𝐹𝑒 = 7.86𝑔𝑐𝑚3 ; 𝐹 = 96484.56


𝑆𝐹𝑒 = 4 × 0.5 × 4 + 4 × 4 × 2 = 40𝑐𝑚2; 𝑡 = 60 × 60 × 24 × 365 = 31536000𝑠


𝐹=

55.845 2 × 10−6 × 31536000

96484.56= 0.0009



ρ𝐹𝑒=

0.00097.86

= 0.00115 𝑐𝑚3

𝑦𝑒𝑎𝑟



=0.00115

40× 10 = 𝟎.𝟎𝟎𝟎𝟐𝟖𝟕𝟓


The amount of corroded Fe will decrease considerable over a year time if the Fe plate is short circuited with Al.

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Figure 08. Evans diagram for a certain steel and aluminum in North Sea water. f) To determine the time in which the Al plate will dissolve, we use the Faraday’s Law and calculate the corrosion rate.

𝑊𝐴𝑙 =𝑀𝑧 × 𝑖𝑐𝑜𝑟𝑟 × 𝑡

𝐹=

26.98 3 × 10−2 × 31536000

96484.56= 32.664


𝑣𝐴𝑙 =𝑊𝐴𝑙

ρ𝐴𝑙=

62.6642.7

= 12.098 𝑐𝑚3

𝑦𝑒𝑎𝑟

Vplate = 4×4×0.5 =8cm3 This means the aluminum will dissolve in 8 months Aluminum can be used as a ‘sacrificial anode ‘because of its high electronegativity and can protect the steel from corrosion.

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