Question

A voltaic cell is constructed from an Ni2+(aq)−Ni(s)Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s)Ag+(aq)−Ag(s) half-cell. The initial concentration of Ni2+(aq)Ni2+(aq) in the Ni2+−NiNi2+−Ni half-cell is [Ni2+]=[Ni2+]= 1.40×10−2 MM . The initial cell voltage is +1.12 VV . Calculate the standard emf of this voltaic cell.

Answer 1

+1.03 V

Step-by-step explanation:

The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).

The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:

Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V

Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V

The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.

emf = E°reduces - E°oxides

emf = 0.80 - (-0.23)

emf = +1.03 V