Question

A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell. The initial concentration of Ni2+(aq) in the Ni2+−Ni half-cell is [Ni2+]= 1.40×10−2 M . The initial cell voltage is +1.12 V .

Answer 1

Step-by-step explanation:

For what I can see, is missing the concentration of [Ag+] in the half-cell. To calculate it:

Niquel half-cell

Oxidation reaction:
`Ni \longrightarrow Ni^(2+)+2 e^-`

`E=E^0 - (R*T)/(n*F)*ln(1/[Ni^(2+)])`

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

`E=-0.23V - (8.314*298)/(2*96500)*ln(1/0.014M)`

`E=-0.285V`

Silver half-cell

Reduction reaction:
`Ag^+ + e^- \longrightarrow Ag`

`E=E^0 - (R*T)/(n*F)*ln(1/[Ag+])`

`E_(cell)=E_(red) - E_(ox)`

`E_(red)=1.12 V + (-0.855V)=0.835V`

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

`0.835V=0.8V - (8.314*298)/(1*96500)*ln(1/[Ag+])`

`[Ag+]=0.26 M`