Question

Ag+(aq) + e– → Ag(s) E° = + 0.80 V
Fe2+(aq)+ + 2e– → Fe(s) E° = – 0.44 V
What is emf of
Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)
(a) 1.16 V
(b) 1.24 V
(c) 2.04 V
(d) -1.16 V

Answer 1

Explanation:

Ag (aq) + e → Ag(s) E° = +0.80 V

Fe2+(aq)+ + 2e → Fe(s) E° = -0.44 V

What is emf of

Fe(s) + 2Ag (aq) → Fe²+(aq) + 2Ag(s)

Answer 2

Given: standard redox potential of silver, E°₁ = + 0.80 V

           standard redox potential of iron, E°₂ = - 0.44 V

To Find: the emf of the cell, E.

Solution:

To calculate E, the formula used:

• Emf of cell = E° cathode - E° anode
• At the cathode, reduction takes place and at the anode oxidation takes place.
• As per the equation given in the question, iron is getting oxidized at the anode, and silver is getting reduced at the cathode.
• ∴ E = E°₁ - E°₂

Applying the above formula :

E = E°₁ - E°₂

  = +0.80 - (-0.44)

  = 0.80 + 0.44

  = +1.24

E = +1.24 V

Hence, the emf of the given cell is 1.24 V (option-b).