Reducing sugars are sometimes characterised by a number Rcu, which is defined as the number of mg of copper reduced by 1 gm of sugar, in which the half reaction of copper is
Cu 2 + 2OH – ⎯→ Cu 2 O + H 2 O
It is sometimes more convenient to determine the reducing power of a carbohydrate by indirect method. In this method 50 mg of the carbohydrate was oxidised by an excess of K 3 [Fe(CN) 6 ]. The Fe(CN) 6 4– formed in this reaction required 10 ml of 0.05 (N) Ce 4+ for reoxidation of Fe(CN) 6 4– to Fe(CN) 6 3– . In this reaction Ce 4+ is reduced to Ce 3+ . The atomic weight of Cu is 63.5. Determine the R cu value of the carbohydrate in nearest possible integers.
Text Solution
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(1) Given Data:
Mass of carbohydrate 
Volume of Ce 4+ solution used 
Normality of Ce 4+ solution 
Half-reaction for copper:
(2) Determine moles of Ce 4+ used:
Normality 
is equivalent to molarity 
for Ce 4+ in this context:
(3) Balanced Redox Reaction:
The reaction between Fe(CN) 6 4– and Ce 4+ is:
For every 1 mole of Fe(CN) 6 4– , 1 mole of Ce 4+ is required.
(4) Determine moles of Fe(CN) 6 4– ,:
(5) Oxidation of carbohydrate:
The carbohydrate reduces Fe(CN) 6 3– , to Fe(CN) 6 4– ,,
Each mole of carbohydrate (as a reducing sugar) is equivalent to 2 moles of Fe(CN) 6 3– ,.
Therefore, moles of carbnhydrate:
Moles of carbohydrate 
(6) Calculate mass of carbohydrate:
Given mass 
.
From the stoichiometry, 
carbohydrate produces 0.00025 mol of Fe(CN) 6 4– ,.
(7) Determine Rcu value:
Rcu is defined as the 
of Cu reduced by 
of sugar.
For the half-reaction:
To find Rcu, we need to determine the moles of Cu 2+ reduced by the carbohydrate. The reduction of Cu 2+ to 
involves 2 moles of Cu 2+ per mole of 
.
(8) Final Calculation:
Moles of Cu 2+ reduced by 
of carbohydrate 
.
Moles of Cu 2+ reduced per 
of carbohydrate:
Since each mole of Cu 2+ corresponds to 
, the Rcu value is:
Thus, the Rcu value of the carbohydrate is approximately 
of Cu per gram of sugar.
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