Home
Class 12
CHEMISTRY
How many hours are needed to deposit the...

How many hours are needed to deposit the metal based on the reaction .
`Fe^(2+)(aq) + 2e^(-) to Fe(s)`. using a current strength of 0.02amp.

Text Solution

Verified by Experts

The correct Answer is:
`268 hr.`
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    AAKASH SERIES|Exercise EXERCISE - 4.1.3|17 Videos
  • ELECTROCHEMISTRY

    AAKASH SERIES|Exercise EXERCISE - 4.1.4|4 Videos
  • ELECTROCHEMISTRY

    AAKASH SERIES|Exercise EXERCISE - 4.1.1|14 Videos
  • DILUTE SOLUTIONS

    AAKASH SERIES|Exercise EXERCISE - 1.2|55 Videos
  • ELEMENTS OF D - BLOCK

    AAKASH SERIES|Exercise EXERCISE - 5.2|33 Videos

Similar Questions

Explore conceptually related problems

The ratio of Fe^(3+) and Fe^(2+) ions in Fe_(0.9)9S_(1.0) is

For the reaction Fe_(2)S_(3)+5O_2 to 2FeSO_4 +SO_(2) the equivalent weight of Fe_(2)S_(3) is-

The standard reduction potentials at 298 K for the following half cell reactions are given below Zn^(2+) (aq) + 2e^(-) to Zn (s)-0.762 Cr^(3+) (aq) + 3e^(-) to Cr (s)-0.740 2H^(+) (aq) + 2e^(-) to H_(2) (g)-0.000 Fe^(3+) (aq) + e^(-) to Fe^(3+) (aq)-0.770 Which one is the strongest reducing agent?

E^(0) for the reaction Fe + Zn^(2+) to Zn + Fe^(2+) is -0.35 V . The given cell reaction is

The standard reduction potentials for the two half-cell reactions are given below : Cd^(2+) (aq) + 2e^(-) to Cd(s) , E^(0) = -0.40 V " "Ag^(+)(aq) + e^(-) to Ag(s) , E^(0) = 0.80 V The standard free energy change for the reaction 2Ag_((aq))^(+) + Cd_((s)) + Cd_((aq))^(2+) is given by

Consider the following E^(0) values E_(Fe^(3+) // Fe^(2+))^(0) = + 0.77 V , E_(Sn^(2+) // Sn)^(0)=-0.14 V Under standard condition the potential for the reaction Sn_((s)) + 2Fe_((aq))^(3+) to 2Fe_((aq))^(2+) + Sn_((aq))^(2+) is

Given the standard oxidation potential Fe overset(+0.4V) to Fe^(2+) (aq) overset(-0.8 V) to Fe^(3+) (aq) , Fe overset(+0.9V) to Fe(OH)_2 overset(0.6V) to Fe(OH)_3 It is easier to oxidize Fe^(2+) to Fe^(3+) in :

A 1.0 M solution of Cd^(+2) is added to excess iron and the system is allowed to reach equilibrium . What is the concentration of Cd^(+2) ? Cd^(2+)(aq) + Fe(s) to Cd(s) + Fe^(2+) (aq) , E^(0) = 0.037