Find EMF of cell Ni-Ni2-Au3-Au if Ni2-Ni- E- -0-25 VAu3-Au- E- - 1-5V

EMF is intensive property hence nbsp; nbsp;E^∘_cell= nbsp;E^∘_cathode E^∘_anode Since Au will act as cathode and Ni will act as anode nbsp; nbsp;E^∘_cell= ...

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Byju's Answer

Standard XII

Chemistry

EMF

Find EMF of c...

Question

Find EMF of cell $N i | N i^{2 +} | | A u^{3 +} | A u$
if
$N i^{2 +} | N i; E^{\circ} = - 0.25 V$
$A u^{3 +} | A u; E^{\circ} = 1.5 V$

1.25 V

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-1.25 V

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1.75 V

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0.83 V

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Solution

The correct option is C 1.75 V
EMF is intensive property hence
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
Since Au will act as cathode and Ni will act as anode
$E_{c e l l}^{\circ} = 1.5 - (- 0.25)$
$E_{c e l l}^{\circ} = 1.75 V$

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Similar questions

Q. Find EMF of cell

N i | N i^{2 +} | | A u^{3 +} | A u

N i^{2 +} | N i; E^{\circ} = - 0.25 V

A u^{3 +} | A u; E^{\circ} = 1.5 V

Q. 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

Q. 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

Q. EMF of cell

N i | N i^{2 +} (1.0 M) | | A u^{3 +} (1.0 M) | A u

(where

E^{\circ}

for

N i^{2 +} | N i

- 0.25 V; E^{\circ}

for

A u | A u^{3 +}

1.50

) is

The EMF of the cell : Ni(s) | Ni²⁺ (1.0M) || Au ³⁺(1.0M) | Au (s) is

[ given E^o_Ni²⁺_/Ni -0.25V; E° _Au³⁺_{/ Au} = +1.50V ]

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Find EMF of cell Ni|Ni2+||Au3+|Au if Ni2+|Ni; E∘= −0.25V Au3+|Au; E∘ = 1.5V

The correct option is C 1.75 VEMF is intensive property hence E∘cell=E∘cathode−E∘anode Since Au will act as cathode and Ni will act as anode E∘cell=1.5−(−0.25) E∘cell=1.75V

Find EMF of cell $N i | N i^{2 +} | | A u^{3 +} | A u$
if
$N i^{2 +} | N i; E^{\circ} = - 0.25 V$
$A u^{3 +} | A u; E^{\circ} = 1.5 V$

The correct option is C 1.75 V
EMF is intensive property hence
$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
Since Au will act as cathode and Ni will act as anode
$E_{c e l l}^{\circ} = 1.5 - (- 0.25)$
$E_{c e l l}^{\circ} = 1.75 V$