Question

$E^{\circ}$ value of $Ni^{2+}/ Ni$ is $-0.25\, V$ and $Ag^+ /Ag$ is $+0.80 \,V$. If a cell is made by taking the two electrodes what is the feasibility of the reaction?

• A. Since $E^{\circ}$ value for the cell will be positive, redox reaction is feasible
• B. Since $E^{\circ}$ value for the cell will be negative, redox reaction is not feasible
• C. $Ni$ cannot reduce $Ag^+$ to $Ag$ hence reaction is | not feasible
• D. $Ag$ can reduce $Ni^{2+}$ to $Ni$ hence reaction is feasible

Answer 1

Answer: (A)

The cell reaction will be
$Ni_{\left(s\right)}+2Ag^{+}_{\left(aq\right)} \to Ni^{2+}_{\left(aq\right)}+2Ag_{\left(s\right)}$
$E^{\circ}_{cell}-E_{cathode}-E_{anode}$
$=0.80-\left(-0.25\right)=+1.05\,V$
$\Delta G^{\circ}=-nFE^{\circ}_{cell}$ As $E^{\circ}_{cell}=+ve$,
$\Delta G^{\circ}=-ve$, hence reaction is feasible.