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Electode potential of Zn^(2+) //Zn is ...

Electode potential of ` Zn^(2+) //Zn` is ` -0 . 7 V` and that of `Cu^(2+) //Cu si = 0. 3 4 V`. The EMF of the cell consttructed between these two elctrodes is .

A

` 2.10 V`

B

` 0.42 V`

C

` -1.1V`

D

` -0. 42 V`

Text Solution

Verified by Experts

The correct Answer is:
A
` E_(cell)^@ =E_("cathode")^@ - D_("anode")^@ = 0. 34 - ( 0.76 ) = 1.1 0V`.
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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`

  • 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

    A
    `-0.42 and 81`
    B
    `1.1 and -213`
    C
    `-1.1 and 213`
    D
    `0.42 and -81`

  • The emf of a galvanic cell with electrode potential of Zn^(2+)//Zn=-0.76 V and that of Cu^(2+)//Cu=+0.34 V is

    A
    `+ 0.34 V`
    B
    `+0.76 V`
    C
    `-1.1 v`
    D
    `+1.1 V`
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