Question

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

Answer 1

Option (B) is correct

Step-by-step explanation:

Oxidation:
`Zn\rightarrow Zn^(2+)+2e^(-)`

Reduction:
`Sn^(2+)+2e^(-)\rightarrow Sn`

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Overall:
`Zn+Sn^(2+)\rightarrow Zn^(2+)+Sn`

Nernst equation for this cell reaction at
`25^(0)\textrm{C}`:

`E_(cell)=(E_(Sn^(2+)\mid Sn)^(0)-E_(Zn^(2+)\mid Zn)^(0))-(0.059)/(n)log([Zn^(2+)])/([Sn^(2+)])`

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

Here, n = 2,
`E_(cell)=0.660V` and
`[Zn^(2+)]=2.5* 10^(-3)M`

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

`0.660V=(-0.136V+0.76V)-(0.059)/(2)log(2.5* 10^(-3)M)/([Sn^(2+)])`

So,
`[Sn^(2+)]=4.2* 10^(-2)M`

As the value "0.059" varies from literature to literature and
`4.2* 10^(-2)M` is most closest to
`3.3* 10^(-2)M` therefore option (B) is correct.