Emf of the cell \[Ni|N{{i}^{2+}}(0.1\,M)||A{{u}^{3+}}\,(1.0\,M)|\,Au\] will be \[(E_{Ni/N{{i}^{2+}}}^{{}^\circ }=0.25,\,E_{Au/A{{u}^{3+}}}^{{}^\circ }=1.5\,V)\]

Emf of the cell \[Ni|N{{i}^{2+}}(0.1\,M)||A{{u}^{3+}}\,(1.0\,M)|\,Au\]

1.	Emf of the cell \[Ni\|N{{i}^{2+}}(0.1\,M)\|\|A{{u}^{3+}}\,(1.0\,M)\|\,Au\] will be \[(E_{Ni/N{{i}^{2+}}}^{{}^\circ }=0.25,\,E_{Au/A{{u}^{3+}}}^{{}^\circ }=1.5\,V)\]
A.	1.75 V
B.	\[+\]1.7795 V
C.	\[+\]0.7795 V
D.	\[-\]1.7795 V
Answer» C. \[+\]0.7795 V

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Emf of the cell \[Ni|N{{i}^{2+}}(0.1\,M)||A{{u}^{3+}}\,(1.0\,M)|\,Au\] will be \[(E_{Ni/N{{i}^{2+}}}^{{}^\circ }=0.25,\,E_{Au/A{{u}^{3+}}}^{{}^\circ }=1.5\,V)\]

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