Home
Class 12
CHEMISTRY
Consult the table of standard electrode ...

Consult the table of standard electrode potential and suggest three substances that can oxidize `Fe^(2+)` ions under suitable conditions.

Text Solution

Verified by Experts

The electrode potential of `Fe^(3+) |Fe^(2+)` is 0.78 V . The substance having electrode potential more than 0.78 V will be reduced and hence act as oxidising agent . For example , (i) acidified `KMnO_(4)` (ii) acidified `Cr_2 O_(7)^(2-)` (iii) acidified `H_(2) O_(2)`.
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    MODERN PUBLICATION|Exercise NCERT FILE NCERT (TEXTBOOK EXERCISES)|27 Videos

  • ELECTROCHEMISTRY

    MODERN PUBLICATION|Exercise NCERT FILE NCERT NCERT EXEMPLAR PROBLEMS (MULTIPLE CHOICE QUESTIONS (TYPE - I))|23 Videos

  • ELECTROCHEMISTRY

    MODERN PUBLICATION|Exercise CONCEPTUAL QUESTIONS|51 Videos

  • D AND F-BLOCK ELEMENTS

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|12 Videos

  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    MODERN PUBLICATION|Exercise COMPETION FILE (Integer Type Numerical Value Type Questions)|5 Videos

Similar Questions

Explore conceptually related problems

The positive value of the standard electrode potential of Cu^(2+)//Cu indicates that.......

On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E^(ϴ) value) (a) Cu+Zn^(2+)toCu^(2+)+Zn (b) Mg+Fe^(2+)toMg^(2+)+Fe (c ) Br_(2)+2Cl^(-)toCl_(2)+2Br^(-) (d) Fe+Cd^(2)toCd+Fe^(2+)

Calculating the emf from standard potentials: A galvanic cell consists of an Al electrode in a 1.0M Al(NO_(3))_(3) solution and an Fe electrode in a 1.0M Fe(NO_(3))_(2) solution. Calculate the standard emf of this electrohemical cell at 25^(@)C . Also write the cell reaction. Strategy: From the table of standard electrode potentials, write the two reduction half reaction and standard electrode potentials for the cell. Change the direction of the half-cell reaction corresponding tol the smaller (or more negative) electrode potential. Multiply the when the half-reactions are added the electrons cancel. The sum of the half-reactions is the cell reaction. Add the electrode potentials to get the cell emf.