Question

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

• A. Since $E^o$ value for the cell will be positive, redox reaction is feasible
• B. Since $E^o$ 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)

Since $E^o$ value for the cell will be positive, redox reaction is feasible

The cell reaction will be

$N{i}_{\left(s\right)}+2A{g}_{\left(aq\right)}^{+}\to N{i}_{\left(aq\right)}^{2+}+2A{g}_{\left(s\right)}$

$E_{cell}^o=E_{cathode}+E_{anode}=0.80-(-0.25)=+1.05V$

$\Delta G=-nF{E}_{cell}^0$

As $E_{cell}^0$ is positive $⟹$ $\Delta G$ is negative

Hence, the reaction is feasible