Question

Adding NaOH to an aqueous solution containing Ni2+ results in the precipitation ofNi(OH)2. The standard potential for the Ni2+/Ni couple is +0.25 V and the solubilityproduct Ksp=[Ni2+][OH− ]2=1.5x10− 16. Calculate the electrode potential at pH=14.

Answer 1

Answer : The electrode potential at pH = 14 is -0.216 V.

Solution : Given,

$E^o_{Ni^{2+}/Ni}$ = +0.25 V

$K_{sp}=[Ni^{2+}][OH^-]^2=1.5\times 10^{-16}$

pH = 14

As we know,

$pH=-log[H^+]=14$

$[H^+]=10^{-14}$

The Solubility product of water is,

$K_w=[H^+][OH^-]=10^{-14}$

$[OH^-]=\frac{K_w}{[H^+]}=\frac{10^{-14}}{10^{-14}}=1$

The reaction is ,

$Ni(OH)_2\rightarrow Ni^{2+}+2OH^-$

The solubility product expression is,

$K_{sp}=[Ni^{2+}][OH^-]^2$

Now put the given values in this expression, we get the $[Ni^{2+}]$ ion.

$1.5\times 10^{-16}=[Ni^{2+}](1)^2$

$[Ni^{2+}]=1.5\times 10^{-16}$

The cell reaction is,

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

Using Nernst equation, $E=E^o-\frac{0.059}{n}logQ$

where,

E = electrode potential of the cell

$E^o$ = standard electrode potential of the cell

n = number of electrons in the cell reaction

Q = reaction quotient

The value of 'n' = 2     (from the cell reaction)

$E=E^o_{Ni^{2+}/Ni}-\frac{0.059}{n}log\frac{1}{[Ni^{2+}]}$

Now put the given values in this formula, we get

$E=+0.25V-\frac{0.059}{2}log\frac{1}{[1.5\times 10^{-16}]}$

E = - 0.216 V

Therefore, the electrode potential at pH = 14 is -0.216 V.