Consider a voltaic cell where the anode half-reaction is zn(s) → zn2+(aq) + 2 e− and the cathode half-reaction is sn2+(aq) + 2 e– → sn(s). what is the concentration of sn2+ if zn2+ is 2.5 × 10−3 m and the cell emf is 0.660 v? the standard reduction potentials are given below

zn+2(aq) + 2 e−→ zn(s) e∘red = −0.76 v

sn2+(aq) + 2 e– →sn(s) e∘red =-0.136 v

a. 9.0*10^-3 m
b. 3..3*10^-2 m
c. 6.9*10^-4 m
d. 7.6*10^-3 m

Option (B) is correct

Explanation:

Oxidation:

Reduction:

---------------------------------------------------------------------------

Overall:

Nernst equation for this cell reaction at :

Where, n is number of electron exchanged and species inside third bracket represents concentrations in molarity.

Here, n = 2, and

So, plug in all the given values into above equation:

So,

As the value "0.059" varies from literature to literature and is most closest to therefore option (B) is correct.