You visited us 1 times! Enjoying our articles? Unlock Full Access!

Primary Cells

E0Ni2+/Ni = -...

Question

$E^{0} (N i^{2 +} / N i) = - 0.25 v o l t, E^{0} (A u^{3 +} / A u) = 1.50 v o l t .$ The standard emf of the voltaic cell. $N i_{(s)} | N i_{(a q)}^{2 +} (1.0 M) | | A u_{(a q)}^{3 +} (1.0 M) | A u_{(s)}$ is:

1.25

volt

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.75

volt

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.75

volt

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.0

volt

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1.75$ volt
$E_{c e l l}^{0} =$ $(E_{N i / {N i}^{2 +}}^{0} + E_{{A u}^{3 +} / A u}^{0})$
$= (0.25 + 1.50) = 1.75 v o l t$

Suggest Corrections

Similar questions

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

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

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

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

Join BYJU'S Learning Program