Question #a3de5

1 Answer
May 27, 2017

The arrangements in Daniel cell
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Anode reaction

#Zn(s)->Zn^(2+)(aq)+2e^-#

Cathode reaction

#Cu^(2+)(aq)+2e^(-) ->Cu(s)#

Overall reaction

#Zn(s)+Cu^(2+)(aq)->Cu(s)+Zn^(2+)(aq)#

The cell notation

#Zn(s)|Zn^(2+) ||Cu^(2+)|Cu(s)#

The Cell Emf by Nernst equation

#E_(cell)=E_(cell)^@-(RT)/(nF)xx([Zn^(2+)])/([Cu^(2+)])......[1]#

If we add sodium sulphide #(Na_2S)# to cupper sulphate in daniel cell, it interacts with #Cu^(2+)# in solution and lowers the concentration of #Cu^(2+)# by precipitating it as insoluble #CuS(s)# as per the following equation.

#Cu^(2+)(aq)+Na_2S->CuS(s)darr+2Na^+#

As a result the ratio #([Zn^(2+)])/([Cu^(2+)])# in equation [1] increases and this decreases the #E_(cell)# to a great extent.