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A student wishes to determine the chloride ion concentration in a water sample at 25 Ā°C using a galvanic cell constructed with a graphite electrode and a half-cell of AgCl(s) + eā» ā†’ Ag(s) + Clā»(aq) EĀ°red = 0.2223 V And a copper electrode with 0.500 M CuĀ²āŗ as the second half cell CuĀ²āŗ(aq) + 2 eā» ā†’ Cu(s) EĀ°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?

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The question is incomplete, the complete question is;

And a copper electrode with 0.500 M CuĀ²āŗ as the second half cell

CuĀ²āŗ(aq) + 2 eā» ā†’ Cu(s) EĀ°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) EĀ°cell = EĀ°cathode - EĀ°anode

EĀ°cell = 0.337 V - 0.2223 V

EĀ°cell = 0.1147 V

b) Ī”GĀ°cell = āˆ’nFEĀ°cell

Where n=2 and F = 96500C

Ī”GĀ°cell =-(2 * 96500 * Ā 0.1147 )

Ī”GĀ°cell =-22,137.1 J or -22.1371 KJ

c) 2Ag(s) + 2Clā»(aq) + CuĀ²āŗ(aq) -----> 2AgCl(s) + Cu(s)

d) From Nernst Equation;

E= EĀ°cell - 0.0592/n Ā log Q

0.0925 Ā = Ā 0.1147 - 0.0592/2 log 1/[0.500] [Clā»]^2

0.0925 - 0.1147 = - 0.0592/2 log 1/[0.500] [Clā»]^2

-0.0222 = -0.0296 log 1/[0.500] [Clā»]^2

-0.0222/-0.0296 = log 1/[0.500] [Clā»]^2

0.75 = log 1/[0.500] [Clā»]^2

Antilog (0.75) = 1/[0.500] [Clā»]^2

5.6234 * 0.500 = Ā [Clā»]^2

[Clā»] = āˆš2.8117

[Clā»] = 1.68 M

A galvanic cell is a voltaic cell that produces electrical energy from the oxidation-reduction process. The standard cell potential of the cell is 0.1147 V.

What is standard cell potential?

The standard cell potential is a difference between the electrode potential of the cathode and the anode of the cell.

The balanced overall reaction of the cell is given as,

[tex]\rm 2Ag(s) + 2Cl^{-}(aq) + Cu^{2+}(aq) \rightarrow 2AgCl(s) + Cu(s)[/tex]

The standard cell potential is calculated as:

[tex]\begin{aligned} \rm E^{\circ}cell &= \rm E^{\circ}cathode - E^{\circ} anode\\\\&= 0.337 - 0.2223 \\\\&= 0.1147\;\rm V\end{aligned}[/tex]

Therefore, 0.1147 V is the standard cell potential for the cell.

Learn more about standard cell potential here:

https://brainly.com/question/17153273

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