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E(Cu^(2+)//Cu)^(@)=0.43V E(Cu^(+)//Cu)...

`E_(Cu^(2+)//Cu)^(@)`=0.43V
`E_(Cu^(+)//Cu)^(@)=0.55V`
`E_(Cu^(2+)//Cu^(+))^(@)`=

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AI Generated Solution

To solve for \( E^\circ_{Cu^{2+} // Cu^+} \), we can use the relationship between the standard electrode potentials of the half-reactions involved. The equation we will use is derived from the concept of combining half-cell reactions: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] In this case, we can rearrange the equation to find the unknown potential: ...
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E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

Given E_(Ag^(+)//Ag)^(@)=+0.80 V, E_(Cu^(2+)//Cu)^(@)=+0.34 V, E_(Fe^(3+)//Fe^(2+))^(@)=+0.76 V, E_(Ce^(4+)//Ce^(3+))^(@)=+1.60 V Which of the following statements is not correct ?

Given E_(Ag^(+)//Ag)^(@)=0.80V , E_(Mg^(2+)//Mg)^(@)=-2.37V , E_(Cu^(2+)//Cu)^(@)=0.79 E_(Hg^(2+)//Hg)^(@)=1.71V Which of the following statements is/are correct?

Four different solution containing 1M each of Au^(+3), Cu^(+2), Ag^(+), Li^(+) are being electrolysed by using inert electrodes. In how many samples, metal ions would be deposited at cathode? ["Given :" E_(Ag^+//Ag)^(0) = 0.8, E_(Au^(+3)//Au)^(0) = 1.00 V E_(Cu^(+2)//Cu)^(0) = 0.34 V, E_(Li^(+)//Li)^(0)= -3.03 V ]

Find the equilibrium constant for the reaction Cu^(2+)+ 1n^(2+) Cu^(+) + 1n^(3+) Given that E_(CU^(2+)//CU^(+))^(@) = 0.15V, E_(In^(2+)//1n^(+)) = -0.4V E_(1n^(3+)//1n^(+))^(@) = - 0.42V

Given: E_(Zn^(+2)//Zn)^(@) =- 0.76V E_(Cu^(+2)//Cu)^(@) = +0.34V K_(f)[Cu(NH_(3))_(4)]^(2+) = 4 xx 10^(11) (2.303R)/(F) = 2 xx 10^(-4) At what concentration of Cu^(+2) emf of the cell will be zero (at 298K) and concentration of Zn^(+2) is remains same:

Given that E_(Cu^(2+)//Cu)^(0)=0.337 and E_(Cu^(+)//Cu^(2+))^(0)=-0.153V . then calculate E_(Cu^(+)//Cu)^(0)

Cu^(+) ion is not stable in aqueous solution because because of dispropotionation reaction. E^(@) value of disproportionation of Cu^(+) is [E_(Cu^(2+)//Cu^(+))^(@)=+ 0.15 V, E_(Cu^(2+)//Cu)^(@)=0.34 V]

Knowledge Check

  • Given E_(Ag^(+)//Ag)^(@)=+0.80 V, E_(Cu^(2+)//Cu)^(@)=+0.34 V, E_(Fe^(3+)//Fe^(2+))^(@)=+0.76 V, E_(Ce^(4+)//Ce^(3+))^(@)=+1.60 V Which of the following statements is not correct ?

    A
    `Fe^(3+)` does not oxidise `Ce^(3+)`.
    B
    Cu reduces `Ag^(+) ` to Ag.
    C
    Ag will reduce `Cu^(2+)` to Cu.
    D
    `Fe^(3+)` oxidises Cu to `Cu^(2+)`

  • Cu^(+) ion is not stable in aqueous solution because because of dispropotionation reaction. E^(@) value of disproportionation of Cu^(+) is [E_(Cu^(2+)//Cu^(+))^(@)=+ 0.15 V, E_(Cu^(2+)//Cu)^(@)=0.34 V]

    A
    second ionisation entyhalpy of copper is less than the first ionisation enthalpy
    B
    large value of second ionisation enthalpy of copper is compensated by much more negative hydration energy of `Cu_((aq))^(2+)`
    C
    hydration energy of `Cu_((aq))^(2+)` is much more negative than that of `Cu_((aq))^(2+)`
    D
    many copper (I) compounds are unstable in aqueous solution and undergo disproportionation rection.