GS-2 Physical Sciences Question 20

[sabha]

Hi, I'm having trouble understanding how the bicarbonate (HCO3-) neutralizes the product of the cathode and how that makes it an inhibitor? What would a cathodic promoter do? I'm sorry..this may be a really simple concept, but I'm having trouble understanding it. Can someone help me please?

[admin]

To begin with, it is not necessarily expected that you should know the term cathodic inhibitor but it is expected that you can deduce the answer from the information provided.

The passage shows that the hydroxide anion is an important feature in causing rust. If you add something which would neutralize hydroxide then you would inhibit the formation of rust. That substance (bicarbonate), therefore, could be referred to as an inhibitor. A promoter, presumably, would result in more hydroxide ions creating more rust.

The next question becomes: is the hydroxide anion formed at the cathode or the anode? Since the hydroxide is being formed as a result of a reduction (Equation II), that terminal is referred to as the cathode ("GERC"= Gain Electrons Reduction at the Cathode which is true for electrochemical/galvanic cells).

[anuradharo8690]

i am still a bit unclear about this...according to le chatelier's principle, wouldn't neutralizing the OH- (thus decreasing the concentration) lead more OH- being formed at the cathode therefore making it a promoter?

i think the part that is confusing me is that the inhibitor part is intended for the rusting process instead of what may be happening at the cathode itself?

clarification would be really appreciated!

[admin]

Yes, it is true that Le Chatelier suggests that as OH- is being removed, the equation shifts to replace the OH-. However, this does not change the fact that as the OH- is being produced, it is being neutralized. Even with Le Chatelier, the concentration of OH- never increases. Le Chatelier tries to return the equation to equilibrium but, even that, will not be successful as long as there is enough bicarb to neutralize OH- as it is being produced thus inhibiting the production of rust.

[anglinak5255]

i chose anoidic inhibitor, since it is preventing the oxidation of Fe to Fe+2. was that wrong thinking process, can you please explain??
Thank you!

[mcat_premed3832]

Please try going through the explanation in the problem and the one above and then specify exactly what about the explanation does not make sense. This exercise will link some important facts. Please understand, you are not supposed to memorize "anodic or cathodic inhibitor" but you must be open to new information presented in a passage or question and then respond based on previous knowledge and reasoning skills. Going back over the explanations step by step will help take you there.

[geoffn045751]

[jellywing_2058]

It is not exactly correct to think of only equation I. The question is with regards to OH- and its direct action in involvement as is in equations II and III and not directly seen in equation I.

As such, the correct term would be cathodic inhibitor (answer B) as the explanation specifies fairly well the rationale or argument.

Moreover, the hydroxide is formed in equation II at the cathode. Subsequently, the HCO3- neutralizes OH- and thus it can be referred to as a cathodic inhibitor. It inhibits the OH- reaction to rusting at the cathode or where it is formed. Thus, the correct answer is answer choice B, cathodic inhibitor.

[charineel4141]

I still don't understand why the answer isn't A.

Being a cathodic inhibitor implies that the reduction REACTION is being inhibited. Even when bicarbonate is there, the oxygen is still being successfully reduced to OH-, so the reaction at the cathode is not being inhibited at all.

The RUSTING process is what's being inhibited. The rusting process is the conversion of iron metal to iron hydroxide, and this is an oxidation reaction, which would happen at the anode. Since the bicarbonate prevents OH- from reacting with Fe, it is preventing the OXIDATION reaction. This would mean that it is an anodic inhibitor.

The bicarbonate does not affect the reduction of oxygen to hydroxide, so it does not inhibit the reaction at the cathode.