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Suppose that the diffusion current in a polarogram for reduction of Cd2+at a mercury electrode is14渭础If the solution containsof 25mLof0.50mMCd2+what percentage ofCd2+is reduced in the 3.4 min required to scan from-0.6to-1.2V?

Short Answer

Expert verified

The percentage of cadmium ion which can be reduced in polarogram is 0.12%.

Step by step solution

01

Define Molarity.

At any point in an electric circuit the charge is the product of current I and time.

q = l . t

The number of moles of electronsreactreacts in a reaction with a given timeis

Reacted moles=l.tnF

02

 Determine the Molarity of Cadmium.

Given,

l=1410-6At=3.4min=204sec

Number of moles of electrons is given as

role="math" =(1410-6A)(204sec)96485C/mol=2.910-8mol

The reduced cadmium ion is calculated as =2.910-8mol2=1.4810-8mol

The 25mL of cadmium ions with has =250.510-31000=1.2510-5moles

Reduced \% of cadmium ions=1.4810-8moles1.2510-5moles100

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