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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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Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100  

The Actual Yield is given in the question as 21.2 g of NaCl.  However, in order to find the theoretical yield, you have to write a balanced equation and use the mole ratio to calculate the mass of NaCl that would be produced.

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Moles of CuCl = Mass of CuCl ÷ Molar Mass of CuCl
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Now, Mass of NaCl = Moles of NaCl × Molar Mass of NaCl
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Now, since Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100  

⇒ Percentage Yield of NaCl = (21.2g ÷ 18.32g) × 100  
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NOTE: Typically, the percentage yield of a reaction is less than 100%, however in a case where the mass of the substance is weighed with impurities, then that mass may be in excess of 100% as seen here.
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