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

A. 1.4 V

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

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.