The EMF of the cell - Ni s - Ni 2-1-0 M Au 3-1-0 M - Au s is- given -E 0 Ni 2- Ni -0-25 V - E - Au 3- - Au -1-50 V -A- 1-25 VB- 2-00 VC- 1-75 VD- -1-25 V

The correct option is C 1.75V E°cell = E°cathode E°anode = E° Au3+ / Au E° Ni2+ / Ni =1.50 + 0.25 = 1.75 V ...

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The EMF of th...

Question

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 ]

1.25V

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2.00 V

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-1.25V

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1.75V

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Similar questions

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

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

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

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

E^{\circ}

N i^{2 +} | N i

- 0.25 V; E^{\circ}

A u | A u^{3 +}

1.50

E^{0} (N i^{2 +} / N i) = - 0.25 v o l t, E^{0} (A u^{3 +} / A u) = 1.50 v o l t .

N i_{(s)} | N i_{(a q)}^{2 +} (1.0 M) | | A u_{(a q)}^{3 +} (1.0 M) | A u_{(s)}

E^{o} ({N i}^{2 +} / N i) = - 0.25 v o l t, E^{o} ({A u}^{3 +} / A u) = 1.50 v o l t

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

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