Question

The reduction potential of following half cell reactions are Mgt2 + 2 e-- Mg; E° = -2.36 V Ni+2 + 2 e-_ Ni; E° = -0.25 V The emf of cell is 0 2.11 V 0 2.61 V 0 2.25 V O 1.09 V the end of cell is​

Answer 1

Answer:

2.11 V

Explanation:

The half cell which has has least reduction potential , acts as the anode so the reaction undergoes oxidation instead . For a spontaneous galvanic/voltaic cell

(Anode) Mg - Mg 2+ + 2e [E° = 2.36]

(cathode) Ni2+ + 2e - Ni [E° = -0.25]

adding

cell reaction - Mg + Ni2+ - Mg2+ + Ni

E cell = 2.36-0.25 = 2.11

Answer 2

Given: E°cell (Mg²⁺/Mg) = -2.36V

          E°cell( Ni²⁺/Ni) = -0.25V

To find: The emf of cell

Solution:

• The cell which undergoes oxidation, and has the least reduction potential is known as an anode. It lies towards the left side of the cell.
• The cell which undergoes reduction, and has the most reduction potential is known as cathode. It lies towards the right side of the cell.
• Electromotive force (EMF) is defined as the electric potential produced by either an electrochemical cell. EMF is commonly used as a short form for electromotive force.

At anode, Mg → Mg²⁺ + 2e⁻ [E° = 2.36]

At cathode, Ni²⁺ + 2e⁻ Ni [ E° = -0.25]

Now adding the both equation we gets -

Mg²⁺ + Ni → Mg²⁺ + Ni

E° cell = 2.36 - 0.25

          = 2.11 V

Therefore, the emf of a cell is 2.11V and the correct option to the question is (a).