Electrochemistry 1 Question 46
45. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $10^{-6} \mathrm{M}$ hydrogen ions. The emf of the cell is $0.118 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. Calculate the concentration of hydrogen ions at the positive electrode.
$(1988,2 \mathrm{M})$
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Solution:
- $\because$ Emf $=0.118 \mathrm{~V}>0$, it is galvanic cell and anode is negative electrode :
At anode : $\quad \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}\left(10^{-6} \mathrm{M}\right)+2 e^{-}$
At cathode : $2 \mathrm{H}^{+}(x)+2 e^{-} \longrightarrow \mathrm{H}_{2}$
Cell reaction : $\quad \mathrm{H}^{+}(x) \longrightarrow \mathrm{H}^{+}\left(10^{-6} \mathrm{M}\right)$
$\mathrm{Emf}=0.118 \mathrm{~V}=0-0.0592 \log \frac{10^{-6}}{x} \Rightarrow x=10^{-4} \mathrm{M}$
