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

Solution : Given,

= +0.25 V

pH = 14

As we know,

The Solubility product of water is,

The reaction is ,

The solubility product expression is,

Now put the given values in this expression, we get the ion.

The cell reaction is,

Using Nernst equation,

where,

E = electrode potential of the cell

= 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)

Now put the given values in this formula, we get

E = - 0.216 V

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

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