I already posted this question 2rd time now someone didn\'t answer the whole thi

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I already posted this question 2rd time now someone didn't answer the whole thing this is a 2rd refund please just answer all parts:

You are required to answer all parts of the question. Our professor gave it as you see it as a single question, so to save to getting a refund and posting again just please.

I know the rule is 1 question at a time in material science at our university this is a single question!!

And correctly clear writting please.

it would be a pain for me to separate them, since I have a fair bit of them.

And finally please do not write on a piece of paper then take a single photo with your mobile phone its going to be blured we can't see it use a photo copier. otherwise take multiple close up photos if mobile is all you have.

8. One end of an iron wire is immersed in an electrolyte of o.02 tre of F ions and the other end in an electrolyte of O.OO5 mole/litre of Fe ions. The two electrodes are separated by a porous wall. By showing your calculations and explaining your reasoning, determine: i) Which end of the wire will corrode marks) ii) The potential difference between the two ends of the wire when it is just immersed in the electrolytes. (3 marks) Nernst equation: E- Eo 0.059 log CCD Standard electrode potential: F Fe 2-+ 2e: E 0.447 b) Explain what is meant by passivation in the context of metal corrosion. Specify three types of anodic reaction that can occur during corrosion of a metal (6 marks) and explain which type causes passivation and why. c) Under aggressive corrosion conditions it is estimated that the maximum corrosion current in a galvanised steel sheet will be G io" A m Estimate the thickness of the galvanised layer needed to give a rust-free life of at least 5 years OG marks) Data for zinc: density 7.13 Mg m Atomic mass 65.38 g mol Itm: The corrosion rate in g s s given by w where F 9650 C mol

Explanation / Answer

8a) 1. The end of wire immrsed in more dilute solution get corroded. that is 0..05 mole/lit

b) For 0.005 mole/lit EFe+ = Eo + 0.0296 log C ion= -0.447+0.0296 log 0.005= -0.515V

For 0.02 mole/lit EFe+ = -(Eo + 0.0296 log C ion)= -(-0.447+0.0296 log 0.02)= 0.497V

Ecell = -0.515 + 0.497= -0.018V

b. Passivation is the process of treating or coating a metal in order to reduce the chemical reactivity of its surface. In stainless steel, passivation means removing the free iron from the surface of the metal using an acid solution to prevent rust.

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