Question

Zinc granules are added in excess to 500 ml of 1.0 M nickel nitrate solution at 25℃ until the equilibrium is reached. If the standard reduction potentials of \[Z{{n}^{2+}}/Zn\] and \[N{{i}^{2+}}/Ni\]are -0.75 and -0.24 volt respectively, find out the concentration of \[N{{i}^{2+}}\] ions in solution at equilibrium.

Answer 1

Hint: The reaction between the zinc granules and the nickel nitrate solution is an example of redox reaction., where zinc is oxidized while nickel gets reduced. The chemical reaction is given as:
\[Z{{n}_{(s)}}+Ni{{(N{{O}_{3}})}_{2}}_{(aq)}\to Zn{{(N{{O}_{3}})}_{2}}_{(aq)}+N{{i}_{(s)}}\]

Complete step by step solution:

We have been given in the question that the:
Volume of nickel nitrate solution = 500 ml
Standard reduction potentials of \[Z{{n}^{2+}}/Zn\] = -0.75 V
Standard reduction potentials of \[N{{i}^{2+}}/Ni\] = -0.24 V

The net redox reaction can be given as:
\[\begin{align}
  & Zn+N{{i}^{2+}}\to Z{{n}^{2+}}+Ni \\
 & initially\,\,\,\,\,\,\,\,\,\,\,500\,\,\,\,\,\,\,\,\,\,\,0 \\
 & at\,equilibrium\,\,\,\,\,\,\,\,\,\,(500-a) \\
\end{align}\]
\[{{E}_{cell}}={{E}_{OP(Zn/Z{{n}^{2+}})}}+{{E}_{RP(N{{i}^{2+}}/Ni)}}\]
\[{{E}_{cell}}={{E}^{\circ }}_{OP(Zn/Z{{n}^{2+}})}+{{E}^{\circ }}_{RP(N{{i}^{2+}}/Ni)}+\dfrac{0.059}{2}\log \dfrac{\left[ N{{i}^{2+}} \right]}{\left[ Z{{n}^{2+}} \right]}\]
We know that at equilibrium, \[{{E}_{cell}}=0\]
Therefore, we can write:
\[0={{E}^{\circ }}_{OP(Zn/Z{{n}^{2+}})}+{{E}^{\circ }}_{RP(N{{i}^{2+}}/Ni)}+\dfrac{0.059}{2}\log \dfrac{\left[ N{{i}^{2+}} \right]}{\left[ Z{{n}^{2+}} \right]}\]
On rearranging, we get:
\[{{E}^{\circ }}_{OP(Zn/Z{{n}^{2+}})}+{{E}^{\circ }}_{RP(N{{i}^{2+}}/Ni)}=-\dfrac{0.059}{2}\log \dfrac{\left[ N{{i}^{2+}} \right]}{\left[ Z{{n}^{2+}} \right]}\]
\[{{E}^{\circ }}_{OP(Zn/Z{{n}^{2+}})}+{{E}^{\circ }}_{RP(N{{i}^{2+}}/Ni)}=-\dfrac{0.059}{2}\log \dfrac{\left[ N{{i}^{2+}} \right]}{\left[ Z{{n}^{2+}} \right]}\]
\[\dfrac{0.51\times 2}{0.059}=\log \dfrac{\left[ N{{i}^{2+}} \right]}{\left[ Z{{n}^{2+}} \right]}\]
Taking antilog on both sides,
\[\begin{align}
  & \dfrac{\left[ N{{i}^{2+}} \right]}{\left[ Z{{n}^{2+}} \right]}=\text{antilog}\left( \dfrac{0.51\times 2}{0.059} \right) \\
 & =5.15\times {{10}^{-18}} \\
\end{align}\]
Now, from the ratio of concentration of can be written as:
\[\dfrac{a}{(500-a)}=5.15\times {{10}^{-18}}\]
Since’ a’ is very small we can write, \[500-a\approx 500\].
\[a=500\times 5.15\times {{10}^{-18}}\]
The concentration of nickel at equilibrium can be calculated as
\[\begin{align}
  & \left[ N{{i}^{2+}} \right]=\dfrac{a}{V}=\dfrac{500\times 5.15\times {{10}^{-18}}}{500} \\
 & =5.15\times {{10}^{-18}}\,M \\
\end{align}\]

Therefore, the concentration of nickel at equilibrium \[5.15\times {{10}^{-18}}\]M.

Note: If the concentrations of the reaction, then there will be another byproduct formed, ammonium nitrate. Hence, it is a concentration dependent reaction and the equation can be given as:
\[4Zn+10HN{{O}_{3}}(conc.)\to 4Zn{{(N{{O}_{3}})}_{2}}+N{{H}_{4}}N{{O}_{3}}+3{{H}_{2}}O\]