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

Explanation:

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

Niquel half-cell

Oxidation reaction: $Ni \longrightarrow Ni^{2+}+2 e^-$

$E=E^0 - \frac{R*T}{n*F}*ln(1/[Ni^{2+}])$

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

$E=-0.23V - \frac{8.314*298}{2*96500}*ln(1/0.014M)$

$E=-0.285V$

Silver half-cell

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

$E=E^0 - \frac{R*T}{n*F}*ln(1/[Ag+])$

$E_{cell}=E_{red} - E_{ox}$

$E_{red}=1.12 V + (-0.855V)=0.835V$

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

$0.835V=0.8V - \frac{8.314*298}{1*96500}*ln(1/[Ag+])$

$[Ag+]=0.26 M$