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Here a reaction is given:

$$\ce{CrO4^2- -> Cr}$$

I don't know the full reaction, but I made some estimates. One estimate was that on balancing charges we get two electrons on right-hand side.

I will consider two different methods that I used for finding $n$-factor:

  1. $n$-factor is the number of electrons participating in a reaction;
  2. $n$-factor in the total change in oxidation state per molecule.

But when I solved by the first method using charge balancing, I got $n$-factor as $1,$ but when I solved by the second method I got $6.$

I am confused in this about how to get the correct $n$-factor. Please check my work and help me to reach out the answer.

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    $\begingroup$ The corresponding half-reaction is :$$\ce{CrO4^{2-} + 8 H^+ + 6 e- -> Cr + 4 H2O}$$ This reaction is used to deposit a bright layer of chromium metal on steel (bumpers) by electrolysis. $\endgroup$
    – Maurice
    Jul 5 at 14:46
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    $\begingroup$ $n$-Factor is an obsolete concept that is essentially useless for over a century. $\endgroup$
    – andselisk
    Jul 5 at 15:07
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    $\begingroup$ @andselisk But it is extremely important for solving problems in physical chemistry for the exams which I am preparing for... $\endgroup$ Jul 5 at 15:11
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    $\begingroup$ @MridulKumar In the context of electrochemistry and Nernst equation in particular, you need the the number of transferred electrons, which is not the same as a more broad and ephemeral term "$n$-factor". If you'll try to use the concept of $n$-factors for, say, multistep electrode process, you'll fail big time. My advice is to avoid erroneous terminology and obsolete concepts, as well as the textbooks that use them. $\endgroup$
    – andselisk
    Jul 5 at 15:31
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    $\begingroup$ The $n$ in Nernst equation is numerically equal to the n-factor. Your teacher is right but this is bad way to connect Nernst equation with equivalents. $\endgroup$
    – M. Farooq
    Jul 5 at 22:54
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Please tell your teachers that the concept of chemical equivalents is obsolete. It was taught 50 years ago. Hope your teachers can communicate to the public exam authorities!

$$\ce{CrO4^2- -> Cr}$$

If it were simple molarity, you would take the experimental weight of dichromate (Salt) and convert that into moles by dividing the mass by formula weight of dichromate salt.

Now you are interested in converting the molar concentration into normality $N$.

In the case of redox reactions, molarity can be converted into normality by dividing it by an integer $n$ and this is your n-factor.

Check how many electrons are transferred: Cr (VI) to Cr (0). So $n$=6. So whatever molarity you have, divide it by 6 to obtain the corresponding normality.

Just follow your second method. Your first point

$n$-factor is the number of electrons participating in a reaction;

does not mean anything.

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