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:

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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.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

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