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I performed an electrolysis of an aqueous sodium carbonate solution with a platinum electrode and a chromium-vanadium-steel-alloy electrode. The solution turned yellow at the chromium-vanadium-alloy electrode (I suspect that it is sodium metavanadate (NaVO3))

Then I also performed an electrolysis in an aqueous sodium carbonate solution with a platinum electrode and an inox electrode (the inox electrode is composed of iron, chromium,...) . A green compound (insoluble) was formed and when I later added acetic acid, a brown compound (I think Iron(III) acetate ) was formed.

But in both cases there were no chromium compounds formed (at least of what I know) So why didn't the chromium react (why didn't the chromium get oxidized)?

More information: Voltage was around 13 Volts, the inox electrode was stainless steel (see http://en.wikipedia.org/wiki/Stainless_steel) and has according to wikipedia a minimum of 10.5% chromium content by mass. I read online that a regular chromium-vanadium-steel-alloy has a vanadium content of approximately 0.181% and a chromium content approximately 1.00%. (however, I'm not sure about these last two figures) What is true is that there wasn't a lot of yellow sodium metavanadate after an electrolysis of around 20 minutes. ) The electrolysis solution contained sodium carbonate(Na2CO3) and water(H2O) the electrolysis was performed at normal conditions (1atm - 25°C)

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  • $\begingroup$ First determine what was the actual result what compounds formed - just assuming the result isn't getting anywhere- also the exact condition under which the experiments were done could help- voltage, actual alloys used, chemistry of solutions. $\endgroup$
    – user2617804
    Commented Dec 19, 2013 at 13:04
  • $\begingroup$ See my edit for more detailed information - I'm also sure that vanadium ions are in the solution because when I reduced the ions from the 2nd liquid (by using zinc) it matched the color changes of vanadium +V,+IV,+III,+II (yellow->blue->green->lilac) $\endgroup$
    – user2117
    Commented Dec 19, 2013 at 14:16
  • $\begingroup$ That doesn't prove that there aren't also Chromium ions in solution. The most vibrant coloured ions are the ones you see. I saw that in wikipedia but inox is just a common name finding the actual alloy has given me no success. $\endgroup$
    – user2617804
    Commented Dec 20, 2013 at 7:10
  • $\begingroup$ the inox-alloy was "18/10 stainless steel" $\endgroup$
    – user2117
    Commented Dec 21, 2013 at 13:23

2 Answers 2

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Some of the ways you aren't at standard condition. Metal solution so the activity of the metals are at their concentration in the alloys The solutions have far below 1mol/L of their respective metal ions The oxygen that you are reducing in basic conditions is probably about 258 micromol per litre- whereas it should be 1 mol/L. The potential might be too high- its hard to tell the solution resistance is an effect.

If you were at standard condition with low voltage then the first experiment makes perfect sense-
From wikipedia full list of electropotentials V2+ + 2 e− V(s) −1.13

Cr3+ + 3 e− Cr(s) −0.74 Vanadium is preferred.

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  • $\begingroup$ So what voltage would you advise to use for the first experiment? $\endgroup$
    – user2117
    Commented Dec 21, 2013 at 13:27
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    $\begingroup$ I'm not the person to ask about practical electrolysis. Either repeat it many times starting from low voltage or contact an organisation like NACE (nace.org) if you are in the Americas or ACA (corrosion.com.au) if you are in Australasia to link you to someone who knows. $\endgroup$
    – user2617804
    Commented Dec 21, 2013 at 14:08
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I agree as Vanadium is preferred, no other metal would be subject to corrosion until all the preferential metal is consumed.

Research the target of a sacrificial electrode.

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