Question

Write the reaction of the electrochemical cell formed with the help of $E_{N{i}^{2+}|Ni}^0=-0.23V$ and $E_{A{g}^{+}|Ag}^0=0.83$ and give symbolic representation.

Answer 1

Metals having a high negative value (low positive value) of reduction potential $\left(E^0\right)$ are good reducing agents, hence get oxidized readily. While metals having positive value (low negative value) of reduction potential are poor reducing agents (better oxidizing agents) and get reduced easily.

$E_{N{i}^{+2}|Ni}^0$ is negative and $E_{A{g}^{+}|Ag}^0$ is positive, therefore in this cell $Ni$ will be oxidised to $N{i}^{+2}$ and $A{g}^{+}$ will get reduced to $Ag$.

Reduction half-reaction at cathode, $2A{g}^{+}+2e^{-}\to 2Ag;E_{A{g}^{+}|Ag}^0=+0.83V$

Oxidation half-reaction at anode, $Ni\to N{i}^{+2}+2e^{-};E_{Ni|N{i}^{+2}}^0=+0.23V$

Cell reaction,

$2A{g}^{+}+Ni\to N{i}^{+2}+Ag;E^0=0.83+0.23=+1.06V$