Consider the cell, P t mid H2(g, 1 atm) H+(a q, 1 M) | F e3+(a q) F e2+(a q) mid P t(s) Given that Eo F e3+ / F e2+=0.771 V, the ratio of concentration of F e(a q)2+ to F e(a q)3+ is, when the cell potential is 0.830 V.

Question

Consider the cell, P t mid H2(g, 1 atm) H+(a q, 1 M) | F e3+(a q) F e2+(a q) mid P t(s) Given that Eo F e3+ / F e2+=0.771 V, the ratio of concentration of F e(a q)2+ to F e(a q)3+ is, when the cell potential is 0.830 V. (A) 0.101 (B) 0.924 (C) 0.120 (D) None of these

Tardigrade Admin · Accepted Answer

The half-cell reactions of the given cell are as
At anode

\frac{1}{2} H_{2} \to H^{+} + e^{-}; E_{1}^{o} = - 0.00 V

At cathode

F e^{3 +} + e^{-} \to F e^{2 +}; E_{?}^{o} = 0.771 V

Net reaction

F e^{3 +} + \frac{1}{2} H_{2} \to F e^{2 +} + H^{+}; E_{cell}^{o} = 0.771 V - 0.00 V = 0.771 V

From Nernst

E_{cell} = E_{cell}^{o} = - \frac{0.0591}{n} lo g \frac{[ F e ^{2 +} ] [ H ^{+} ]}{[ F e ^{3 +} ] + p H _{2}^{1/2}}

0.830 = 0.771 - \frac{0.0591}{1} lo g [\frac{F e ^{2 +}}{F e ^{3 +}}] - \frac{0.059}{0.0591} = lo g \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]}

∴ \frac{[ F e ^{2 +} ]}{[ F e ^{3 +} ]} =

antilog

(- 0.998) = 0.101

Q. Consider the cell, Pt∣H2​(g,1atm)H+(aq,1M)∥Fe3+(aq) Fe2+(aq)∣Pt(s) Given that EoFe3+/Fe2+​=0.771V, the ratio of concentration of Fe(aq)2+​ to Fe(aq)3+​ is, when the cell potential is 0.830V.

0.101

0.924

0.120

None of these

Q. Consider the cell,
$Pt ∣ H_{2} (g, 1 a t m) H^{+} (a q, 1 M) ∥ F e^{3 +} (a q)$ $F e^{2 +} (a q) ∣ Pt (s)$
Given that $E^{o}_{F e^{3 +} / F e^{2 +}} = 0.771 V$ , the ratio of concentration of $F e_{(a q)}^{2 +}$ to $F e_{(a q)}^{3 +}$ is,
when the cell potential is $0.830 V$ .