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oksian1 [2.3K]
3 years ago
14

A student wishes to determine the chloride ion concentration in a water sample at 25 °C using a galvanic cell constructed with a

graphite electrode and a half-cell of AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq) E°red = 0.2223 V And a copper electrode with 0.500 M Cu²⁺ as the second half cell Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red= 0.337 V The measured cell potential when the water sample was placed into the silver side of the cell was 0.0925 V. What is the standard cell potential for this cell in V?
Chemistry
1 answer:
soldi70 [24.7K]3 years ago
4 0

The question is incomplete, the complete question is;

And a copper electrode with 0.500 M Cu²⁺ as the second half cell

Cu²⁺(aq) + 2 e⁻ → Cu(s) E°red= 0.337 V

The measured cell potential when the water sample was placed into the silver side of the cell was 0.0925 V.

A- What is the standard cell potential for this cell in V?

B- What is the value of the standard free energy (in kJ) for this reaction?

C- Write the balanced equation for the overall reaction in acidic solution?

D- And the measured cell potential is 0.0925, what is the concentration of chloride ions in the solution?

Answer:

See Explanation

Explanation:

a) E°cell = E°cathode - E°anode

E°cell = 0.337 V - 0.2223 V

E°cell = 0.1147 V

b) ΔG°cell = −nFE°cell

Where n=2 and F = 96500C

ΔG°cell =-(2 * 96500 *  0.1147 )

ΔG°cell =-22,137.1 J or -22.1371 KJ

c) 2Ag(s) + 2Cl⁻(aq) + Cu²⁺(aq) -----> 2AgCl(s) + Cu(s)

d) From Nernst Equation;

E= E°cell - 0.0592/n  log Q

0.0925  =  0.1147 - 0.0592/2 log 1/[0.500] [Cl⁻]^2

0.0925 - 0.1147 = - 0.0592/2 log 1/[0.500] [Cl⁻]^2

-0.0222 = -0.0296 log 1/[0.500] [Cl⁻]^2

-0.0222/-0.0296 = log 1/[0.500] [Cl⁻]^2

0.75 = log 1/[0.500] [Cl⁻]^2

Antilog (0.75) = 1/[0.500] [Cl⁻]^2

5.6234 * 0.500 =  [Cl⁻]^2

[Cl⁻] = √2.8117

[Cl⁻] = 1.68 M

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Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

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CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

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According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

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C

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Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

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