NEET PYQ Electrochemistry | Chapter-Wise Solutions

(3)

pH = 7 for water.

$- \log [H^{+}] = 7 \Rightarrow [H^{+}] = 10^{- 7}$

$2 H_{(a q)}^{+} + 2 e^{-} ⟶ H_{2 (g)}$

$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{2} \log \frac{p_{H_{2}}}{{[H^{+}]}^{2}}$

$0 = 0 - \frac{0.0591}{2} \log \frac{p_{H_{2}}}{{(10^{- 7})}^{2}}$

$\log \frac{p_{H_{2}}}{{(10^{- 7})}^{2}} = 0 \Rightarrow \frac{p_{H_{2}}}{{(10^{- 7})}^{2}} = 1$ $[∵ \log 1 = 0]$

$p_{H_{2}} = 10^{- 14}$ $atm$

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