Electrolysis . Why does the conductivity of a solution decrease with dilution? . Suggest a way to determine the L ° m value of water.
. The molar conductivity of . mol L – methanoic acid is . S cm mol – .
Calculate its degree of dissociation and dissociation constant. Given l (H + ) = . S cm mol – and l (HCOO – ) = . S cm mol – .
There were no constant current sources available during Faraday’s times. The general practice was to put a coulometer (a standard electrolytic cell) for determining the quantity of electricity passed from the amount of metal (generally silver or copper) deposited or consumed. However, coulometers are now obsolete and we now have constant current ( I ) sources available and the quantity of electricity Q , passed is given by Q = It Q is in coloumbs when I is in ampere and t is in second. The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
For example, in the reaction: Ag + (aq) + e – ® Ag(s) ( . ) One mole of the electron is required for the reduction of one mole of silver ions. We know that charge on one electron is equal to . × – C.
Therefore, the charge on one mole of electrons is equal to: N A × . × – C = . × mol – × . × – C = 96487 C mol – This quantity of electricity is called Faraday and is represented by the symbol F .
For approximate calculations we use 1F ≃ 96500 C mol – . For the electrode reactions: Mg + (l) + 2e – ¾® Mg(s) ( . ) Al + (l) + 3e – ¾® Al(s) ( . ) It is obvious that one mole of Mg + and Al + require mol of electrons (2F) and mol of electrons (3F) respectively.
The charge passed through the electrolytic cell during electrolysis is equal to the product of current in amperes and time in seconds. In commercial production of metals, current as high as , amperes are used that amounts to about . F per second. A solution of CuSO is electrolysed for minutes with a current of .
amperes. What is the mass of copper deposited at the cathode? t = s charge = current × time = . A × s = C According to the reaction: Cu + (aq) + 2e – = Cu(s) We require 2F or × 96487 C to deposit mol or g of Cu.
For C, the mass of Cu deposited = ( g mol – × C)/( × 96487 C mol – ) = . g.