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

Nernst Equation

For the given...

Question

For the given cell, $P t, {{C l}_{2} / H C l}_{(1 a t m, C_{1})} | | {H C l / {C l}_{2}}_{(C_{2}, 1 a t m)}$
the emf will be positive when:

C_{1} > C_{2}

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

C_{2} > C_{1}

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

C_{1} = C_{2}

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

none of these

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

Open in App

Solution

The correct option is B $C_{2} > C_{1}$
Solution:- (B) $C_{2} > C_{1}$
Given cell will be-
$P t, {{C l}_{2} / H C l}_{(1 a t m, C_{1})} | | {H C l / {C l}_{2}}_{(C_{2}, 1 a t m)}$
According to nernst equation-
$E_{c e l l} = E^{\circ} - 0.591 log \frac{C_{1}}{C_{2}}$
$\Rightarrow E_{c e l l} = E^{\circ} - 0.591 log \frac{C_{1}}{C_{2}}$
$\Rightarrow E_{c e l l} = E_{\circ} + 0.0591 log \frac{C_{2}}{C_{1}}$
For $C_{2} > C_{1}$
$\Rightarrow log \frac{C_{2}}{C_{1}} > 0$
$\Rightarrow E_{c e l l} > 0$
Hence for emf to be positive, $C_{2} > C_{1}$ .

Suggest Corrections

Similar questions

C l^{-}

C_{1}

C l_{2}

P_{1}

C l_{2}

P_{1}

C l^{-}

C_{2}

\frac{c_{1} - c_{2}}{\sqrt{1 + m^{2}}}

\frac{|c_{1} - c_{2}|}{\sqrt{1 + m^{2}}}

\frac{c_{2} - c_{1}}{\sqrt{1 + m^{2}}}

|c_{1} - c_{2}|

C_{1}

V_{o}

C_{2}

$A g | A g C l | C l^{-} (C_{2}) | | C l^{-} (C_{1}) | A g C l | A g for this cell Δ G$ is negative if:

v

(x)

v = \sqrt{c_{1} - c_{2} x}

c_{1}

c_{2}

Join BYJU'S Learning Program