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Crystal Field Split in Octahedral Complexes

Pt H2 p1 | H...

Question

$P t (H_{2}) (p_{1}) | H C l (0.1 M) | (C l_{2}) (p_{2})$ Pt cell reaction will be endergonic if:

p_{1} = p_{2}

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p_{1} > p_{2}

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p_{2} > p_{1}

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p_{1} = p_{2} = 1

atm

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Solution

The correct option is C $p_{2} > p_{1}$
in endergonic
$Δ G > 0$ and $E_{c} e l l < 0$
so $P t (H_{2})_{p} 1 | H C l (0.1 m) | (C l_{2})_{p} 2$
$(C l_{2}) p_{1} \to C l$
$C l^{-} \to {C l_{2}}_{(p_{2})}$
$Δ E_{c e l l} = \frac{- 0.0591}{n} \times l o g \frac{p_{2}}{p_{1}}$
so answer is
$P_{2} > P_{1}$

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