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.