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Catalysis

Pt H2 p1 | H...

Question

$P t (H_{2}) (p_{1}) | H^{+} (1 M) (H_{2}) (p_{2})$ , Pt cell reaction will be exergonic if:

A

p_{1} = p_{2}

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B

p_{1} > p_{2}

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C

p_{2} > p_{1}

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D

p_{1} = 1 a t m

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Solution

The correct option is C $p_{1} > p_{2}$
For reaction to be exergonic
$△ C_{1} = - v e \Rightarrow E_{c e l l} = + v e E_{c e l l} = 0 - \frac{0.0591}{1} l o g \frac{{[H}^{+}]_{A} \times (P_{1 / 2})_{C}^{1 / 2}}{(P_{1 / 2})_{A} ({[H}^{+}]_{C}} = 0.0591 l o g \frac{P_{2}}{P_{1}}$
For $E_{c e l l} = + v e P_{1} > P_{2}$

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Similar questions

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

p_{1}

p_{2}

are pressure) cell reaction will be spontaneous if:

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}

are pressure) cell reaction will be spontaneous if:

Q. The cell reaction for the given cell is

P t (H_{2} (P_{1})) | p H = 2 | | p H = 3 | P t (H_{2} (P_{2}))

P_{1} = 1 a t m

P_{2} = 1 a t m

.

P t

C l_{2}

P_{1}

H C l

0.1

C l_{2}

P_{2}

P t

, cell reaction will be spontaneous if:

p_{1}

p_{2}

are two odd prime numbers such that

p_{1} > p_{2}

p_{1}^{2} - p_{2}^{2}

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