Electrochemistry Question 4
Question 4 - 25 January - Shift 2
$Pt(s) H_2(g)( 1 bar ) \mid H^{+} (aq )((1M) ) \mid M^{3+}(aq), M^{+}(aq) \mid Pt(s)$
The $E _{\text{cell }}$ for the given cell is $0.1115 V$ at $298 K$
when $\frac{[M^{+}(aq)]}{[M^{3+}(aq)]}=10^{a}$
The value of $a$ is
Given : $E^{\theta}_ {M^{3+} / M^{+}}=0.2 V$
$\frac{2.303 RT}{F}=0.059 V$
Show Answer
Answer: (3)
Solution:
Overall reaction :-
$H _{2(g)}+M _{(aq)}^{3+} \longrightarrow M _{(aq)}^{+}+2 H _{(aq)}^{+}$
$E _{\text{Cell }}=E _{\text{Cathode }}^{o}-E _{\text{anode }}^{o}-\frac{0.059}{2} \log \frac{[M^{+}] \times 1^{2}}{[M^{+3}] 1}$
$0.1115=0.2-\frac{0.059}{2} \log \frac{[M^{+}]}{[M^{+3}]}$
$ 3=\log \frac{[M^{+}]}{[M^{+3}]} $
$\therefore a=3$
