Discussion Forum

1)

For the following electrochemical cell at 298 K,

$Pt(s) | H_{2}(g,1bar) | H^{+}(aq,1M)$

                                    $|| M^{4+}(aq), M^{2+}(aq)| Pt (s)$

$E_{cell}$=0.092 V

$when \frac{[M^{2+}(aq)]}{[M^{4+}(aq)]}=10^{x}$

Given:  $E^{0}_{M^{4+/}M^{2^{+}}}=0.15V:$

$2.303 \frac{RT}{F}=0.059 V$

 The value of x is

Answer:

Option D

Explanation:

Oxidation at anode

$H_{2}(g) \rightarrow 2H^{+}(aq) +2e^{-}; E^{0}_{SHE}=0.00V$

Reduction at cathode

$M^{4+}(aq) +2e^{-}\rightarrow M^{2+}(aq):$

$E^{0}_{M^{4+}/M^{2+}}=0.151V$

$Net: M^{4+}(aq)+H_{2}(g)\rightarrow M^{2+}(aq)+2H^{+}(aq):$

$K= \frac{[M^{2+}][H^{+}]^{2}}{[M^{4+}]PH_{2}}(E^{0}_{cell}=0.151 V)$

$K= \frac{[M^{2+}]}{[M^{4+}]}$

$E_{cell}=E^{0}_{cell}-\frac{0.059}{2}\log K$

$0.092= 0.151_{}-\frac{0.059}{2}log \frac{[M^{2+}]}{[M^{4+}]}$

$0.059= \frac{0.059}{2}log 10^{x}$

$\therefore \log 10^{x}=2$

$\therefore x=2$