Daniel cell operates under nonstandard state conditions. If the equation of the cell reaction is multiplied by 2 then:

E

and

E^{o}

remain unchanged

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E

is doubled

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n

remains unchanged in Nernst equation

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Q

is halved in Nernst equation

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Solution

The correct option is B $E$ and $E^{o}$ remain unchanged
Nernst's equation states $E_{c e l l} = E_{c e l l}^{0} + \frac{R T}{n F} l n [M^{+}]$
Where $E$ and $E^{0}$ of cell are independent of volume/quantity of reacting species in electrolyte, they are intunsic properties of cell. Hence, on doubling cell equation of Daniel cell.
$Z n ∣ ∣ {Z n}^{2 +} ∣ ∣ ∣ ∣ {C u}^{2 +} ∣ ∣ C u$ ,
its e.m.f $(E)$ and standard e.m.f $(E^{0})$ remain unchanged where $E^{0}$ stays at $1.09 V$ .

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Similar questions

Z n (s) + C u_{(a q)}^{2 +} \to Z n_{(a q)}^{2 +} + C u_{(s)}

M_{(a q)}^{n +} + n e^{-} \to M_{(s)}

E_{M^{n +} / M} = E^{o}

M^{n +} / M - \frac{2.303 R T}{n F} l o g \frac{[M]}{[M^{n +}]}

E_{(c e l l)}^{o}

(K_{c})

E = E^{o} - \frac{R T}{n F} l n Q

Q = K_{c}

E = E^{\circ} - \frac{2.303 R T}{n F} {log}_{10} Q

Q

K_{C}

E

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