Question

What is the cell potential of an electrochemical cell that has the half -reactions shown below? Ni^ 2+ +2e Ni A| A|^ 3+ +3e^ -

A. 1.4V
B.-1.93 V
C. 1.43v
D. 261 V

Answer 1

A. 1.4 V

Step-by-step explanation:

We are given the half reactions;

Ni²⁺(aq) + 2e → Ni(s)

Al(s) → Al³⁺(aq) + 3e

We are required to determine the cell potential of an electrochemical cell with the above half-reactions.

E°cell = E(red) - E(ox)

From the above reaction;

Ni²⁺ underwent reduction(gain of electrons) to form Ni

Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺

The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V

Therefore;

E°cell = -0.25 V - (-1.66 V)

= -0.250V + 1.66 V

= + 1.41 V

= + 1.4 V

Therefore, the cell potential will be +1.4 V