Question

$E^{\circ }$ value of $N{i}^{2+}/Ni$ is $-0.25V$ and $A{g}^{+}/Ag$ is $+0.80V$. 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 $A{g}^{+}$ to $Ag$ hence reaction is | not feasible
• D. $Ag$ can reduce $N{i}^{2+}$ to $Ni$ hence reaction is feasible

Answer 1

Answer: (A)

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