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The E^(@) value of Zn is -0.76 V while t...

The `E^(@)` value of Zn is -0.76 V while that of Cu is +0.34 V. Do these values help in locating the relative positions of the electrodes in the electrochemical series ?

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Yes, they do help.
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E^(@) of Cu is +0.34 V while that of Zn is -0.736 V. Explain.

E^(@) of Cu is + 0.34 V while that of Zn is – 0.76 V. Explain.

Knowledge Check

  • Electrode potential of Zn^(2+)//Zn is -0.76V and that of Cu^(2+)//Cu is +0.34V. The EMF of the cell constructued between these two electrodes is

    A
    1.10V
    B
    0.42V
    C
    `-1.1V`
    D
    `-0.42V`

  • Given that E^(@)(Zn) = -0.76V and E^(@)(Cu) = 0.34V . Calculate the emf of the cell with cell reaction

    A
    `-42 V`
    B
    `1.42 V`
    C
    `-1.10 V`
    D
    `+1.10 V`

  • E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

    A
    `Cu^(2+)` Oxidizes Fe
    B
    `Fe^(2+)` oxidizes Cu
    C
    Cu Reduces `Fe^(2+)`
    D
    Fe reduces `Cu^(2+)`

    • Similar Questions

    Explore conceptually related problems

    E^(c-) of Cu is + 0.34 V while that of Zn is - 0.76V . Explain.

    E^(Theta) of Cu is +0.34V while that of Zn is -0.76 V. Explain.

    The standard electrode potentials at 298 k are given below E_(Zn^(2+)//Zn)^(@)=-0.76 V, E_(Fe^(2+)//Fe)^(@)=-0.44V E_(H^(+)//H_(2))^(@)=0.0V and E_(Cu^(2+)//Cu)^(@)=+0.34V Which of the two electrodes should be combined to form a cell having highest EMF? Identify the cahode and the anode and write the cell reaction also mention the direction flow of electrons in the external as well as the internal circuit

    The standard electrode potentials at 298 k are given below E_(Zn^(2+)//Zn)^(@)=-0.76 V, E_(Fe^(2+)//Fe)^(@)=-0.44V E_(H^(+)//H_(2))^(@)=0.0V and E_(Cu^(2+)//Cu)^(@)=+0.34V Which of the two electrodes should be combined to form a cell having highest EMF? Identify the cahode and the anode and write the cell reaction also mention the direction flow of electrons in the external as well as the internal circuit

    The standard electrode potential of Zn^(2+) // Zn is -0.76V and that of Ca^(2+) // Cu is 0.34V . The emf(V) and the free energy change (kJ "mol"^(-1)) , respectively , for a Daniel cell will be

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