Electrochemistry Question 12
Question 12 - 01 February - Shift 1
At what $pH$, given half cell $MnO_4^{-}(0.1 M) \mid Mn^{2+}$
$(0.001 M)$ will have electrode potential of 1.282
$V$ ? (Nearest Integer)
Given $E _{MnO_4^{-} / Mn^{2+}}^{o}=1.54 V, \frac{2.303 R T}{F}=0.059 V$
Show Answer
Answer: (3)
Solution:
$MnO_4^{-}+8 H^{+}+5 e^{-} \rightarrow Mn^{2+}+4 H_2 O$
$E=E^{\circ}-\frac{0.059}{5} \log \frac{[Mn^{2+}]}{[MnO_4^{-}][H^{+}]^{8}}$
$1.282=1.54-\frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times[H^{+}]^{8}}$
$\frac{0.258 \times 5}{0.059}=\log \frac{10^{-2}}{[H^{+}]^{8}}$
$\Rightarrow \quad 21.86=-2+8 pH$
$\therefore \quad pH=2.98$
$\simeq 3$
