ElectrochemistryHard
Question
The EMF of the cell: Zn|Zn2+ (0.01 M)||Fe2+ (0.001 M)|Fe at 298 K is 0.2905 V, then the value of equilibrium constant for the cell reaction is
Options
A.$e^{\frac{0.32}{0.0295}}$
B.$10^{\frac{0.32}{0.0295}}$
C.$10^{\frac{0.26}{0.0295}}$
D.$10^{\frac{0.32}{0.0591}}$
Solution
Cell reaction: $Zn + Fe^{2 +} \rightleftharpoons Zn^{2 +} + Fe$
$E_{cell} = E_{cell}^{o} - \frac{0.059}{2}.\log\frac{\left\lbrack Zn^{2 +} \right\rbrack}{\left\lbrack Fe^{2 +} \right\rbrack} $$${\text{Or, 0.2905 = }\frac{0.059}{2}.\log K_{eq} - \frac{0.059}{2}.\log\frac{0.01}{0.001} }{\therefore K_{eq} = 10^{\left( \frac{0.32}{0.0295} \right)}}$$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
Pt (where p1 and p2 are pressures) cell reaction will be spontaneous if :...The electrical conductivity of a solution serves as a means of determining the end point in a chemical reaction, involve...The standard EMF for the cell reaction: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) is 1.10 volts at 25°C. The EMF of the cell...Calculate the solubility product of Co2[Fe(CN)6] in water at 25oC.Given, conductivity of saturated solutions of Co2[Fe(C...A current of 9.95 amp following for 10 minutes, deposits 3 gm of a metal. Equivalent weight of the metal is...