Question

A concentration cell consists of two sn/sn2+ half-cells. the electrolyte in compartment a is 0.24 m sn(no3)2. the electrolyte in b is 0.87 m sn(no3)2. which half-cell houses the cathode? what is the voltage of the cell?

Answer 1

A:-      sn(s) =>  Sn +2(0.24 M) + 2e-
B:-     Sn +2 (0.87 M) +2e-  =>   Sn(s) 

solution will become more concentrated and solution B become less concentrated 

Sn(s)+ Sn +2(0.87 ) ----> Sn(s) + Sn  +2(0.24)
E  =   Eo  -   0.0592 / 2 * log   [   (0.24 / 0.87 ) ]
E  = 0.0   -   0.0592 / 2   *   log ( 0.275) 
( n=2 two electrons are transferred)

E =  -0.0296 *  ( - 0.560) 
E =  0.0165 volts