EMF of cell Ni - Ni2-1-0M - - Au3-1-0 M - Au where E- for Ni2- - Ni is -0-25 V-E- for Au - Au3- is 1-50 is

The correct option is C +1.75 V E_cell^o=E^o_Au^3+/Au E^o_Ni^2+/Ni=1.50 ( 0.25)=1.75 V . ...

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EMF

EMF of cell ...

Question

EMF of cell $N i | N i^{2 +} (1.0 M) | | A u^{3 +} (1.0 M) | A u$ (where $E^{\circ}$ for $N i^{2 +} | N i$ is $- 0.25 V; E^{\circ}$ for $A u | A u^{3 +}$ is $1.50$ ) is

+ 1.25

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- 1.75

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+ 1.75

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+ 4.0

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Similar questions

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

E^{o}

N i^{2 +} / N i

- 0.25

V

E^{o}

A u^{3 +} / A u

0.150 V

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

E^{0}

N i^{2 +} | N i

- 0.25

E^{0}

A u^{+ 2} | A u

1.50

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 ]

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

E^{\circ} = - 0.25 V

N i^{2 +} | N i; E^{\circ} = 1.5 V

A u^{3 +} | A u

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

E^{\circ} = - 0.25 V

N i^{2 +} | N i; E^{\circ} = 1.5 V

A u^{3 +} | A u

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