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Calculate the standard electrode potenti...

Calculate the standard electrode potential of `Ni^(2+)`/Ni electrode if emf of the cell `Ni_((s)) |Ni^(2+) (0.01M)| |Cu^(2)| Cu _((s))(0.1M)` is `0.059 V.` `[Given : E_(Cu^(2+)//Cu)^(@) =+0.34V]`

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To calculate the standard electrode potential of the Ni²⁺/Ni electrode, we can follow these steps: ### Step 1: Understand the Cell Reaction The cell reaction involves two half-reactions: - At the cathode: Cu²⁺ + 2e⁻ → Cu (s) - At the anode: Ni (s) → Ni²⁺ + 2e⁻ ### Step 2: Identify the Given Values ...
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Calculate the standard electrode potential of the Ni^(2+)//Ni electrode if the cell potential of the cell Ni|Ni^(2+)(0.01M)||Cu^(2+)(0.1M)|Cu is 0.59 V. Given E_(Cu^(2+)//Cu)^(@)=+0.34V .

Calculate the standard electrode potential of the Ni^(2+)//Ni electrode , if the cell potential potential of the cell, Ni//N^(2+)(0.01 M)//Cu is 0.59" V ". "Given" E_(Cu^(2+)//Cu)^(@)=+0.34 " V "

Knowledge Check

  • If the standard electrode poten tial of Cu^(2+)//Cu electrode is 0.34V. What is the electrode potential of 0.01 M concentration of Cu^(2+) ?

    A
    0.399V
    B
    0.281V
    C
    0.222V
    D
    0.176V

  • The standard potential of Cu | Cu^(2+) electrode is -0.34 V . It corresponds to the reaction

    A
    `Cu to Cu^(2+) + 2 e^(-)`
    B
    `Cu^(2+) + 2e^(-) to Cu`
    C
    `Cu^(+) to Cu^(2+) + e^(-)`
    D
    `Cu to Cu^(+) - e^(-)`

  • If the standard electrode potential of Cu^(2+)//Cu electrode is 0.34 V. What is the electrode potential of 0.01 M aqueous solution of Cu^(2+) at 298K?

    A
    `0.399` V
    B
    `0.281` V
    C
    `0.222` V
    D
    `0.176` V

    • Similar Questions

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    Calculate the standard reduction potential of Ag^(+)|Ag electrode when the cell potential for the cell , Cu (s) |Cu^(2+) (1 M) || Ag^(+) (1 M) Ag is 0.46 V . Given that Cu^(2+) |Cu = 0.34 V

    Calculate the standard reduction potential of the following half cell S^(2-)|CuS|Cu Given: K_(sp)(CuS)=10^(-35)andE_(Cu^(2+)//Cu)^(@)=0.34V

    (a) A cell is prepared by dipping a zinc rod in 1M zinc sulphate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential given : E^(@)Zn_(2+1//Zn) = -0.76V, E^(@)A_(g+//)A_(g) = +0.80V What is the effect of increase in concentration of Zn^(2+) " on the " E_(cell) ? (b) Write the products of electrolysis of aqueous solution of NaCI with platinum electrodes. (c) Calculate e.m.f. of the following cell at 298 K: "Ni(s)"//"Ni"^(2+)(0.01M)////"Cu"^(2+)(0.1M)//"Cu(s)" ["Given"E_(Ni2+//Ni)^(@) = -0.025 V E_(Cu2+//Cu)^(@) = +0.34V] Write the overall cell reaction.

    Emf of the cell Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M) Au will be E_(Ni//Ni(2+))^@ = 0.5=25. E_(Au//Au^(3+))^@ = 1.5 V .

    If standard electrode potenial of Cu^(2+)//Cu is 0.34V then potential of Cu dipped in 0.1 M solution of CuSO_(4) will be

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