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.

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.