Why is it said that reduction by sugars is much more efficient in basic media than in acidic media as in the following quote taken from Satyanaryana’s Biochemistry, 4th edition page 16?

In the laboratory, many tests are employed to identify the reducing action of sugars. These include Benedict’s test, Fehling’s test, Barfoed’s test etc. The reduction is much more efficient in the alkaline medium than in the acidic medium. (sic emphasis)

Is it something to do with the reagents used because many of them are just basic solutions, but if the reagent happens to be an acidic solution then will be that statement valid?

  • 2
    $\begingroup$ Fun fact about the quote: All three of those tests are the same reaction with mildly different conditions. But note that Tollens’ reagent requires the same conditions. $\endgroup$
    – Jan
    Nov 8, 2016 at 15:51
  • 1
    $\begingroup$ On a side note, Barfoed actually mentions slightly acidic conditions in his original article DOI. $\endgroup$ Nov 8, 2016 at 16:41

1 Answer 1


This is based in the underlying redox rection. If we take e.g. mannose and attempt to oxidise that, the (unbalanced) half-reaction we need is the following:

$$\ce{C6H12O6 -> C6H10O6 + 2e-}\tag{Ox1}$$

To balance it, we either need to add protons to the right-hand side (acidic medium) or negative charges to the left-hand side. Since we are asked about alkaline solutions, let’s do the latter:

$$\ce{C6H12O6 + 2 OH- -> C6H10O6 + 2 e- + 2 H2O}\tag{Ox2}$$

As you can see, the oxidation of sugars consumes hydroxide ions. If you increase the concentration of hydroxide, that will typically increase the reaction rate.

Of course, one should always check what the reduced compound is. In your case, it is always a complexated copper(II) ion and is reduced by the following equation:

$$\ce{2Cu^2+ + 2 e- + 2 OH- -> Cu2O v + H2O}\tag{Red}$$

Even here, if we set up the half-reaction, we notice that we are consuming hydroxide. Thus, the overall redox reaction is:

$$\ce{C6H12O6 + 2 Cu^2+ + 4 OH- -> C6H10O6 + Cu2O v + 3 H2O}\tag{Redox}$$

If a reaction consumes hydroxide ions, it is enhanced by basic conditions. Likewise, if a redox reaction consumes protons, it is enhanced by acidic conditions.

  • $\begingroup$ so implicit in this is that if I increase acidity then the reaction will proceed backwards, since to balance electrons we need to add protons on product hand side? Also in what form does the hydrogen is removed in the unbalanced reaction? $\endgroup$
    – JM97
    Nov 8, 2016 at 16:05
  • $\begingroup$ @JM97 I learnt to set up redox half-reactions in the following way: 1) identify the redox pair and the electron difference; add them. 2) balance the charge using $\ce{H3O+}$ ($\ce{H+}$) or $\ce{OH-}$, depending on conditions. 3) balance the mass, usually be checking the number of oxygens or hydrogens. The hydrogens in $(\mathrm{Ox1})$ do not participate in redox chemistry, hence they are ignored until we arrive at the third step which is after charge balancing. Thus, they are initially missing. $\endgroup$
    – Jan
    Nov 8, 2016 at 16:08
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    $\begingroup$ Also, the reaction won’t necessarily go backwards just because the pH value is reversed to acidic. For some redox reactions (i.e. bromine dis- or comproportionation) it does, but for others the Gibbs free energy difference is too large and simply nothing will happen given suboptimal conditions. $\endgroup$
    – Jan
    Nov 8, 2016 at 16:09
  • $\begingroup$ @vapid I decided to go from the aldehyde form to a linear mannonic acid form, please do not assume I wanted to go to any other form. Adding something like $\ce{+ [O]}$ was not appreciated and not desired for the first equation. Finally, I ask you please never to change British spellings to American spellings, especially in my posts. $\endgroup$
    – Jan
    Nov 9, 2016 at 19:39
  • $\begingroup$ What is this $\ce{C6H10O6}$ compound then? If you wanted a linear mannonic acid as you wrote, then the formula is $\ce{C6H12O7}$. $\endgroup$
    – vapid
    Nov 10, 2016 at 8:03

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