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EMF

E o Ni 2+ / ...

Question

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

1.25 v o l t

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2 v o l t

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1.75 v o l t

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4.0 v o l t

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Solution

The correct option is B $1.75 v o l t$
$E_{c e l l}^{o} = E^{o} A u^{3 +} / A u - E^{o} {N i}^{2 +} / N i = 1.50 + 0.25 = 1.75 V E_{c e l l} = 1.75 - \frac{0.0591}{6} l o g \frac{1}{1} E_{c e l l} = 1.75 V$
${R e x}^{n}$ are
$N i ⟶ {N i}^{2 +} + {2 e}^{-}} \times 3 A u^{3 +} + 3 e^{-} ⟶ A u} \times 2___________________3 N i + 2 {A u}^{3 +} ⟶ 3 {N i}^{2 +} + 2 A u$

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Similar questions

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)}

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

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 +} (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

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