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  • Please,please help me - Redox Reaction and Electrochemistry - JEE Main-7

ven : 

Co^{3+}+ e^{-}\rightarrow Co^{2+}+;E^{0}=+1.81V

Pb^{4+}+2 e^{-}\rightarrow Pb{2+}+;E^{0}=+1.67V

Ce^{4+}+ e^{-}\rightarrow Ce{3+}+;E^{0}=+1.61V

Bi ^{3+}+ 3e^{-}\rightarrow Bi;E^{0}=+0.20 V

Oxidising power of the species will increase in the  order: 

 

  • Option 1)

    Bi^{3+}< Ce^{4+}< Pb^{4+}< Co^{3+}

  • Option 2)

    Ce^{4+}< Pb^{4+}< Bi^{3+}< Co^{3+}

  • Option 3)

    Co^{3+}< Ce^{4+}< Bi^{3+}< Pb^{4+}

  • Option 4)

    Co^{3+}< Pb^{4+}< Ce^{4+}< Bi^{3+}

 

Answers (1)

best_answer

 

Electrode Potential -

The potential associated with each electrode is known as electrode potential.

-

 

 

Electro-chemical series -

The standard reduction potential of a large number of electrodes have been measured using standard hydrogen electrode as the reference electrode. These various electrode can be arranged in increasing electrode potential.

-

 

 

 

Standard Electrode Potential Value -

-

 

 

 

 

The greater the Standard Reduction Potential, the more will be its oxidizing power. 

\therefore Correct sequence will be : 

Co^{3+}> Pb^{4+}> Ce^{4+}> Bi^{3+}

 


Option 1)

Bi^{3+}< Ce^{4+}< Pb^{4+}< Co^{3+}

Option 2)

Ce^{4+}< Pb^{4+}< Bi^{3+}< Co^{3+}

Option 3)

Co^{3+}< Ce^{4+}< Bi^{3+}< Pb^{4+}

Option 4)

Co^{3+}< Pb^{4+}< Ce^{4+}< Bi^{3+}

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