Consider the following cellPt - H 2 P 1 atm - H - M 1 H - M 2- H 2 P 2 atm - Ptwhere P 1 and P 2 are pressures- M 1 and M 2 are molarities- What will be the emf of cell at 25- C If P 1- P 2 and M 1 is 50 - higher than M 2 - Repeat the magnitude of emf by multiplying with 10000 - -Take - 2-303 RT - F -0-06 and -log 3-0-48- log 2-0-3-

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Standard XII

Chemistry

Nernst Equation

Consider the ...

Question

Consider the following cell
$P t | H_{2} (P_{1} a t m) | H^{+} (M_{1}) | | H^{+} (M_{2}) | H_{2} (P_{2} a t m) | P t$
where $P_{1}$ and $P_{2}$ are pressures. $M_{1}$ and $M_{2}$ are molarities .
What will be the emf of cell at $25^{\circ} C$ If $P_{1} = P_{2}$ and $M_{1}$ is 50% higher than $M_{2}$ ?
Repeat the magnitude of emf by multiplying with 10000.
[Take : $\frac{2.303 R T}{F} = 0.06 and log 3 = 0.48, log 2 = 0.3]$

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Solution

$P t | H_{2} (P_{1} a t m) | H^{+} (M_{1}) | | H^{+} (M_{2}) | H_{2} (P_{2} a t m) | P t$
at anode $H_{2} \to 2 H^{+} + 2 e^{-}$
$P_{1} (M_{1})$
at cathode $2 H^{+} \to H_{2}$
$(M_{2}) P_{2}$
net cell reaction $2 H^{+} + H_{2} (P_{1}) \to 2 H^{+} + H_{2} (P_{2})$
$(M_{2}) (M_{1})$
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.06}{2} l o g \frac{[H^{+}]^{2} (P_{2})}{(H^{+})^{2} (P_{1})} ∵ P_{1} = P_{2}$
$E_{c e l l}^{0} = 0$
$E_{c e l l} = - \frac{0.06}{2} l o g \frac{[H^{+}]_{M_{1}}^{2}}{[H^{+}]_{M_{2}}^{2}}$
$= - \frac{0.06}{2} l o g (1.5)^{2} = - 0.0108 V$
$∴ 0.0108 \times 10000 = 108$

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