Home
Class 12
CHEMISTRY
If E(Cu^(2+)|Cu)^(@) = 0.34V and E(Cu^(2...

If `E_(Cu^(2+)|Cu)^(@) = 0.34V` and `E_(Cu^(2+)|Cu^(+))^(@)= 0.15 V` then the value for disproportion for `Cu^(+)` is :

A

`(-0.19V)`

B

`(-0.38V)`

C

`(0.94)`

D

`(0.38V)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the value for the disproportionation potential of \( Cu^+ \), we can follow these steps: ### Step 1: Understand the Disproportionation Reaction The disproportionation reaction for \( Cu^+ \) can be represented as: \[ 2 Cu^+ \rightarrow Cu^{2+} + Cu \] In this reaction, \( Cu^+ \) is both oxidized to \( Cu^{2+} \) and reduced to \( Cu \). ### Step 2: Identify the Standard Electrode Potentials From the question, we have the following standard electrode potentials: - \( E^\circ (Cu^{2+}/Cu) = 0.34 \, V \) - \( E^\circ (Cu^{2+}/Cu^+) = 0.15 \, V \) ### Step 3: Write the Half-Reactions We can write the half-reactions based on the given potentials: 1. \( Cu^{2+} + 2e^- \rightarrow Cu \) with \( E^\circ = 0.34 \, V \) (Reduction) 2. \( Cu^{2+} + e^- \rightarrow Cu^+ \) with \( E^\circ = 0.15 \, V \) (Reduction) ### Step 4: Reverse the Second Half-Reaction To find the oxidation potential of \( Cu^+ \) to \( Cu^{2+} \), we reverse the second half-reaction: \[ Cu^+ \rightarrow Cu^{2+} + e^- \quad (E^\circ = -0.15 \, V) \] ### Step 5: Calculate the Overall Cell Potential Now, we can combine the two half-reactions: - The reduction of \( Cu^{2+} \) to \( Cu \) (first half-reaction) - The oxidation of \( Cu^+ \) to \( Cu^{2+} \) (reversed second half-reaction) The overall cell reaction is: \[ Cu^+ + 2e^- \rightarrow Cu + Cu^{2+} \] ### Step 6: Use the Nernst Equation The overall cell potential \( E^\circ_{cell} \) can be calculated using: \[ E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} \] Substituting the values: \[ E^\circ_{cell} = 0.34 \, V + (-0.15 \, V) = 0.34 \, V - 0.15 \, V = 0.19 \, V \] ### Step 7: Calculate the Disproportionation Potential The potential for the disproportionation of \( Cu^+ \) can be calculated as follows: \[ E^\circ_{disproportionation} = E^\circ_{cell} = 0.19 \, V \] ### Final Answer The value for the disproportionation potential for \( Cu^+ \) is: \[ E^\circ_{disproportionation} = 0.19 \, V \] ---

To find the value for the disproportionation potential of \( Cu^+ \), we can follow these steps: ### Step 1: Understand the Disproportionation Reaction The disproportionation reaction for \( Cu^+ \) can be represented as: \[ 2 Cu^+ \rightarrow Cu^{2+} + Cu \] In this reaction, \( Cu^+ \) is both oxidized to \( Cu^{2+} \) and reduced to \( Cu \). ...
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    VMC MODULES ENGLISH|Exercise JEE Main (Archive)|57 Videos
  • ELECTROCHEMISTRY

    VMC MODULES ENGLISH|Exercise JEE Advanced (Archive)|76 Videos
  • ELECTROCHEMISTRY

    VMC MODULES ENGLISH|Exercise Level-1|75 Videos
  • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

    VMC MODULES ENGLISH|Exercise JEE ADVANCE (ARCHIVE)|30 Videos
  • ENVIRONMENTAL CHEMISTRY

    VMC MODULES ENGLISH|Exercise JEE Main (Archive)|39 Videos

Similar Questions

Explore conceptually related problems

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^(+))^(@)=

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

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

E_(Cu^(+2)//Cu)^(@)=0.34V, E_(Zn//Zn^(+2))^(@)=0.76V A cell formed by the conbination of Cu and Zn (a) when CuSO_(4) is added to Cu^(+2) compartment what is the effect on emf of cell (b) when NH_(3) is added to Cu^(+2) compartment what is the effect on emf of cell (c ) When ZnSO_(4) is added to Zn^(+2) compartment is the effect on emf of cell (d) When Zn^(+2) is diluted what is the effect on emf of cell ?

Given: (i) Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V (ii) Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V Electrode potential, E^(@) for the reaction, Cu^(+)+e^(-) rarr Cu , will be

Given that E_(cu^(+2)//cu)^(@)=+0.34 V E_(Mg^(+2)//Mg)^(@)=-2.37V which of the following correct

The standard potential E^(@) for the half reactions are as : Zn rarr Zn^(2+) + 2e^(-), E^(@) = 0.76V Cu rarr Cu^(2+) +2e^(-) , E^(@) = -0.34 V The standard cell voltage for the cell reaction is ? Zn +Cu^(2) rarr Zn ^(2+) +Cu

Consider the following E^(@) values: E_(Li^(+)|Li)^(@)=-3.05V, E_(Cu^(2+)|Cu)^(@)=+0.34V Under similar conditions, the potential for the reaction Cu+2Li^(+)rarrCu^(2+)+2Li , is

The standard reduction potentials of Cu^(2+)|Cu and Cu^(2+)|Cu^(o+) are 0.337V and 0.153V , respectively. The standard electrode potential fo Cu^(o+)|Cu half cell in Volts is

VMC MODULES ENGLISH-ELECTROCHEMISTRY-Level-2
  1. 108 g solution of AgNO(3) is electrolysed using Pt electrodes by pass...

    Text Solution

    |

  2. The standard oxidation potentials, , for the half reactions are as fo...

    Text Solution

    |

  3. If E(Cu^(2+)|Cu)^(@) = 0.34V and E(Cu^(2+)|Cu^(+))^(@)= 0.15 V then th...

    Text Solution

    |

  4. How many Faradays are needed to reduce 1 mole of MnO(4)^(-) to Mn^(2+...

    Text Solution

    |

  5. Two elcetrolytic cells are containing acidified Fe(2)(SO(4))(3) and th...

    Text Solution

    |

  6. Calculate E(cell) for Cr|Cr^(3+)(0.04M)||Cr^(3+)(1M)|Cr:

    Text Solution

    |

  7. Given that K(sp) of CuS=10^(-35) and ECu//Cu^(2+)= (-0.34V). The stan...

    Text Solution

    |

  8. At T (K), the molar conductivity of 0.04 M acetic acid is 7.8 S cm^(2)...

    Text Solution

    |

  9. When water is electrolysed, hydrogen and oxygen gas are produced. If 1...

    Text Solution

    |

  10. A1^(3+) + 3e^(-) to Al(s), E^(@) = -1.66V Cu^(2+)+ 2e^(-) rightarrow ...

    Text Solution

    |

  11. (10Cl^(-)(aq) + 2MnO(4)^(-)(aq) + 16H^(+)(aq) rightarrow 5Cl(2)(g) + 2...

    Text Solution

    |

  12. In an exper iment 0.04F was passed through 400mL of 1 M solution of Na...

    Text Solution

    |

  13. Efficiency of the following cell is 84% A(s)+B(aq)^(2+) + B(s), Delta...

    Text Solution

    |

  14. The ratio of (^^m)/(^^eq) for Ca(3) (po4)2 will be equal to :

    Text Solution

    |

  15. At what pH the potential of Hydrogen electrode will be 0.059 V?

    Text Solution

    |

  16. Number of Faradays required to convert 1 mol of Cr(2)O(7)^(2-) into Cr...

    Text Solution

    |

  17. Charge on 6.24xx10^18 electrons will be (in coulomb):

    Text Solution

    |

  18. Cr(s)| Cr^(3+)||Fe^(2+)|Fe(s)In the above cell, the value of n in the ...

    Text Solution

    |

  19. All the energy released from the reaction Xto Y, Delta(r) G^(@) = -330...

    Text Solution

    |

  20. The molar conductance at infinite dilution for electrolytes BA and CA ...

    Text Solution

    |