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For Cu(OH)2|Cu hall-cell at pH = 12, electrode potential is 0.0455 at 298 K.
For Cu2+ (aq) + 2e- → Cu, E° = 0.34 V. Ksp value of Cu(OH)2 is x x 10-4
What is the value of x?    
    Correct answer is '1'. Can you explain this answer?
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    For Cu(OH)2|Cu hall-cell at pH = 12, electrode potential is 0.0455 at ...
    The given information relates to a Hall cell consisting of Cu(OH)2|Cu electrodes at a pH of 12 and a temperature of 298 K. The electrode potential for this cell is given as 0.0455 V, and the standard electrode potential for the reduction of Cu2+ to Cu is 0.34 V. We are also given the Ksp value of Cu(OH)2 as x × 10^-4 and asked to determine the value of x.

    1. Understanding the Hall Cell:
    A Hall cell consists of two half-cells: an anode and a cathode. In this case, the anode is the Cu(OH)2 electrode, and the cathode is the Cu electrode. The half-reaction occurring at the anode is the oxidation of Cu(OH)2, while the half-reaction occurring at the cathode is the reduction of Cu2+ to Cu.

    2. Determining the Overall Cell Reaction:
    Based on the given information, we can write the overall cell reaction as follows:
    Cu(OH)2(s) → Cu2+(aq) + 2OH^-(aq) + 2e^-

    3. Applying the Nernst Equation:
    The Nernst equation relates the electrode potential of a half-cell to the concentrations of the species involved. In this case, we can use it to relate the electrode potential of the Cu(OH)2 electrode to the concentration of Cu2+.

    The Nernst equation is given by:
    E = E° - (RT/nF) * ln(Q)
    where:
    E is the electrode potential
    E° is the standard electrode potential
    R is the gas constant
    T is the temperature
    n is the number of electrons transferred
    F is Faraday's constant
    Q is the reaction quotient

    4. Calculating the Concentration of Cu2+:
    Since the given pH is 12, we know that the concentration of OH^- ions is 10^-2 M. Using this information, we can write the reaction quotient Q as [Cu2+]/[OH^-]^2.

    5. Substituting Values and Solving for x:
    By substituting the given values into the Nernst equation and solving for E, we get:
    0.0455 V = 0.34 V - (RT/2F) * ln([Cu2+]/[OH^-]^2)

    Using the fact that RT/F at 298 K is approximately 0.0257 V, we can simplify the equation to:
    0.0455 V = 0.34 V - (0.0257 V/2) * ln([Cu2+]/(10^-2 M)^2)

    Simplifying further, we get:
    0.0455 V = 0.34 V - 0.01285 V * ln([Cu2+]/10^-4 M)

    Since ln([Cu2+]/10^-4 M) = ln([Cu2+]) - ln(10^-4 M), we can rewrite the equation as:
    0.0455 V = 0.34 V - 0.01285 V * (ln([Cu2+]) - ln(10^-4 M))

    By rearranging the equation, we obtain:
    ln([Cu2+]) = (0.0455 V - 0.34 V + 0.01285 V * ln(10^-
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    For Cu(OH)2|Cu hall-cell at pH = 12, electrode potential is 0.0455 at 298 K.For Cu2+ (aq) + 2e-→ Cu, E° = 0.34 V. Ksp value of Cu(OH)2 is x x 10-4.What is the value of x? Correct answer is '1'. Can you explain this answer?
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    For Cu(OH)2|Cu hall-cell at pH = 12, electrode potential is 0.0455 at 298 K.For Cu2+ (aq) + 2e-→ Cu, E° = 0.34 V. Ksp value of Cu(OH)2 is x x 10-4.What is the value of x? Correct answer is '1'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about For Cu(OH)2|Cu hall-cell at pH = 12, electrode potential is 0.0455 at 298 K.For Cu2+ (aq) + 2e-→ Cu, E° = 0.34 V. Ksp value of Cu(OH)2 is x x 10-4.What is the value of x? Correct answer is '1'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For Cu(OH)2|Cu hall-cell at pH = 12, electrode potential is 0.0455 at 298 K.For Cu2+ (aq) + 2e-→ Cu, E° = 0.34 V. Ksp value of Cu(OH)2 is x x 10-4.What is the value of x? Correct answer is '1'. Can you explain this answer?.
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