The emf of the cell $Ni\ |\ {Ni}^{2 +}\ (1.0\ M)\ ||\ {Au}^{3 +}\ (1.0\ M)\ |\ Au$ is [$E^{{^\circ}}\ ({Ni}^{2 +}/\ Ni) =$-0.25 V and $E^{{^\circ}}\ ({Au}^{3 +}/\ Au) =$+1.5 V]

Question

Dr. Abubacker · Accepted Answer

(d)
For the cell, $Ni\ |\ {Ni}^{2 +}\ ||\ {Au}^{3 +}|\ Au$
Given, $E_{{Ni}^{2 +}/Ni}^{{^\circ}} = - 0.25V$
$E_{{Au}^{3 +}/Au}^{{^\circ}} = + 1.5\ V$
Here, Ni is anode and Au is cathode.
∴ $E_{cell} = \ E_{C} - \ E_{A}$
= 1.5 – (-0.25)
= 1.5 + 0.25
= 1.75 V

Q. The emf of the cell $Ni\ |\ {Ni}^{2 +}\ (1.0\ M)\ ||\ {Au}^{3 +}\ (1.0\ M)\ |\ Au$ is [$E^{{^\circ}}\ ({Ni}^{2 +}/\ Ni) =$-0.25 V and $E^{{^\circ}}\ ({Au}^{3 +}/\ Au) =$+1.5 V]

Explanation

Reviewed by Dr. Abubacker

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