Can Solution of 1M ZnSO4 be stored in a vessel made up of Nickel? Given: $E_{N{i^{ + 2}}/Ni}^0 = - 0.25V$ $E_{Z{n^{ + 2}}/Zn}^0 = - 0.76$

Question Collected from Telegram.

Solution:

The cell reaction for the process is:

$Zn^{2+} + Ni \rightarrow Zn + Ni^{2+}$

This can be represented using half-reactions as:

Anode (oxidation half-reaction):

$Zn \rightarrow Zn^{2+} + 2e^-$

Cathode (reduction half-reaction):

$Ni^{2+} + 2e^- \rightarrow Ni$

The overall cell reaction is obtained by adding the anode and cathode half-reactions:

$Zn + Ni^{2+} \rightarrow Zn^{2+} + Ni$

Based on the given standard reduction potentials, we can calculate the cell potential for the reaction:

$Zn^{2+} + Ni \rightarrow Zn + Ni^{2+}$

The cell potential can be calculated using the formula:

$E_{cell} = E_{cathode} - E_{anode}$

where $E_{cathode}$ is the reduction potential of the cathode and $E_{anode}$ is the reduction potential of the anode.

Substituting the given values, we get:

$E_{cell} = E_{Ni^{2+}/Ni} - E_{Zn^{2+}/Zn} = (-0.25) - (-0.76) = 0.51V$

Since the cell potential is positive, this indicates that the reaction is spontaneous and the Zinc Shulphate Can be stored in the Vessel of Nickel.