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