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

Answer:

Option (B) is correct

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 $25^{0}\textrm{C}$:

$E_{cell}=(E_{Sn^{2+}\mid Sn}^{0}-E_{Zn^{2+}\mid Zn}^{0})-\frac{0.059}{n}log\frac{[Zn^{2+}]}{[Sn^{2+}]}$

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

Here, n = 2, $E_{cell}=0.660V$ and $[Zn^{2+}]=2.5\times 10^{-3}M$

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

$0.660V=(-0.136V+0.76V)-\frac{0.059}{2}log\frac{2.5\times 10^{-3}M}{[Sn^{2+}]}$

So, $[Sn^{2+}]=4.2\times 10^{-2}M$

As the value "0.059" varies from literature to literature and $4.2\times 10^{-2}M$ is most closest to $3.3\times 10^{-2}M$ therefore option (B) is correct.