The EMF of cell Ni - Ni 2- Au 3- Au is-Given that E -0-25 V for Ni 2- Ni - E -1-5 V for Au 3- AuA- 1-25 VB- -1-25VC- 1-75 VD- 2-0 V

The correct option is C 1.75VE^∘_cell=E^∘_cathode E^∘_anode Cathode here is Au as Au^3+→ Au i.e., reduction at Au electrode and as Ni → Ni^2+ i.e., oxidation a ...

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Standard X

Chemistry

Electrolytic Cell

The EMF of ce...

Question

The EMF of cell $N i | N i^{2 +} | | A u^{3 +} | A u$ is:
Given that $E^{\circ} = - 0.25 V$ for $N i^{2 +} | N i; E^{\circ} = 1.5 V$ for $A u^{3 +} | A u$

1.25V

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-1.25V

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1.75V

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2.0V

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Solution

The correct option is C 1.75V
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
Cathode here is Au as $A u^{3 +} \to A u$
i.e., reduction at Au electrode
and as $N i \to N i^{2 +}$ i.e., oxidation at Ni electrode, it is anode.
So,
$E_{c e l l}^{\circ} = 1.5 - (- 0.25)$
$= 1.75 V$

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The EMF of cell Ni|Ni2+||Au3+|Au is: Given that E∘=−0.25V for Ni2+|Ni;E∘=1.5V for Au3+|Au

The correct option is C 1.75VE∘cell=E∘cathode−E∘anode Cathode here is Au as Au3+→Au i.e., reduction at Au electrode and as Ni→Ni2+ i.e., oxidation at Ni electrode, it is anode. So, E∘cell=1.5−(−0.25) =1.75V

The EMF of cell $N i | N i^{2 +} | | A u^{3 +} | A u$ is:
Given that $E^{\circ} = - 0.25 V$ for $N i^{2 +} | N i; E^{\circ} = 1.5 V$ for $A u^{3 +} | A u$