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
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.
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:
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