What will be the e.m.f of the given cell? $P t | H_{2} (P_{1}) | H^{+} (a q) | H_{2} (P_{2}) | P t$

\frac{R T}{F P} l n \frac{P_{1}}{P_{2}}

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

\frac{R T}{2 F} l n \frac{P_{1}}{P_{2}}

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

\frac{R T}{F} l n \frac{P_{2}}{P_{1}}

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 $\frac{R T}{2 F} l n \frac{P_{1}}{P_{2}}$
$H_{2} (P_{1}) + {2 H}^{+} (1 M) ⟶ {2 H}^{+} (1 M) + H_{2} (P_{2})$
$E_{c e l l} = E_{c e l l}^{0} - \frac{R T}{n F} I n$
$= 0 - \frac{R T}{2 F} I n \frac{P_{2}}{P_{1}}$ [for concentration cell, $E_{c e l l}^{0} = 0$ ]
$= \frac{R T}{2 F} I n \frac{P_{1}}{P_{2}}$

Suggest Corrections

Similar questions

P t | H_{2} (P_{1}) | H^{+} (a q) . | H_{2} (P_{2}) | P t

What will be the emf for the given cell :

$P t | H_{2} (P_{1}) | H^{+} (a q) | H_{2} (P_{2}) | P t$

P t | H_{2} (p_{1}) | | H^{+} (1 M) | | H^{+} (1 M) | | H_{2} (p_{2}) | P t

p_{1}

p_{2}

P t | \frac{H_{2}}{(p_{1})} ∥ \frac{H^{+}}{(1 M)} ∥ \frac{H^{+}}{(1 M)} ∥ \frac{H_{2}}{(p_{2})} | P t

p_{1}

p_{2}

P t

C l_{2}

P_{1}

H C l

0.1

C l_{2}

P_{2}

P t

Join BYJU'S Learning Program