Question

Derive an expression for emf of a concentration cell without transference

Answer 1

For the purpose of understanding first we have to consider two simple cells.

Pt,$H_{2(g)}$,$Hcl_{(a_{1}) }$/$Agcl_{(s)}$,Ag

Pt,$H_{2(g)}$,$Hcl_{(a_{2}) }$/$Agcl_{(s)}$,Ag

1)Here the two electrolytes are not in contact with each other.

2)Activity of $H^{+}$ ions in the two solutions be $(a_{1} )$ and $(a_{2} )$.

3)The cells are comes together in a way that they oppose each other.

so we have to consider the cell

Pt,$H_{2(g)}$,$Hcl_{(a_{1}) }$/$Agcl_{(s)}$,Ag

Cell reactions:

1)$\frac{1}{2}$ $H_{2}$+$Agcl_{(s)}$↔$Ag_{(s)}$+$Hcl_{(a_{1}) }$

2)$\frac{1}{2}$ $H_{2}$+$Agcl_{(s)}$↔$Ag_{(s)}$+$Hcl_{(a_{2}) }$

Then the expression for emf is

$E_{w.o.t}$=$\frac{RT}{F}$ln$\frac{a_{2} }{a_{1} }$