Electrochemistry - Result Question 46
45. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $10^{-6} M$ hydrogen ions. The emf of the cell is $0.118 $ $V$ at $25^{\circ}$ $ C$. Calculate the concentration of hydrogen ions at the positive electrode.
$(1988,2 M)$
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Answer:
Correct Answer: 45. $(10^{-4} M)$
Solution:
- $\because$ Emf $=0.118 V>0$, it is galvanic cell and anode is negative electrode :
At anode : $ H _2(g) \longrightarrow 2 H^{+}\left(10^{-6} M\right)+2 e^{-}$
At cathode : $2 H^{+}(x)+2 e^{-} \longrightarrow H _2$
Cell reaction : $ H^{+}(x) \longrightarrow H^{+}\left(10^{-6} M\right)$
Emf=$0.118 V=0-0.0592 \log \frac{10^{-6}}{x} \Rightarrow x=10^{-4} M$
BETA
