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dusya [7]
3 years ago
11

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
Chemistry
1 answer:
Sonbull [250]3 years ago
8 0

Answer:

Option (B) is correct

Explanation:

Oxidation: Zn\rightarrow Zn^{2+}+2e^{-}

Reduction: Sn^{2+}+2e^{-}\rightarrow Sn

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

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.

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The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

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Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

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where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

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In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

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<h3>Answer:</h3>

1.827 × 10²⁴ molecules H₂S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
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103.4 g H₂S (Sulfuric Acid)

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<u>Step 3: Convert</u>

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  2. Multiply:                                                                                                            \displaystyle 1.82656 \cdot 10^{24} \ molecules \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S

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