For the cell $P t | H_{2} (1 a t m) | H C l (0.01 M) | | H C l (p H = X) | H_{2} (0.01 a t m) | P t$ . Calculate x, when there is no flow of electricity.

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Solution

No flow means zero EMF
$E = E^{\circ} - 0.0591 log \frac{[H^{+}] \times (p H_{2})}{(y)}$
$0 = 0 - 0.0591 log \frac{10^{- 4}}{(y)}$
$0.0591 = log (\frac{10^{- 4}}{y})$
$a n t i l o g 0.06 = \frac{10^{- 4}}{y}$
$0.8 = \frac{10^{- 4}}{y}$
$y = 1.25 \times 10^{- 4}$
$p H (x) = 4 - log 1.25$
$4 - 0.9$
$\approx 3$

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