Question:
The standard emf of a cell involving one electron charge is found to be $0.591\, V$ at $25^{\circ} C$. The equilibrium constant of the reaction is $\left(1\, F =96500\, C\, mol ^{-1}, R=8.314\, J\, K ^{-1} mol ^{-1}\right)$
The standard emf of a cell involving one electron charge is found to be $0.591\, V$ at $25^{\circ} C$. The equilibrium constant of the reaction is $\left(1\, F =96500\, C\, mol ^{-1}, R=8.314\, J\, K ^{-1} mol ^{-1}\right)$
Updated On: Jul 28, 2022
- $ 1.0\times 10^{1}$
- $ 1.0\times 10^{30}$
- $ 1.0\times 10^{10}$
- $ 1.0\times 10^{5}$
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
At $298\, K$
$E^{\circ} =\frac{0.0591}{n} \log K_{c}$
$0.591 =\frac{0.0591}{1} \log K_{c}$
$\log K_{c} =\frac{0.5910}{0.0591}$
$\log K_{c} =10$
$K_{c} =10^{10}$
Was this answer helpful?
0
0
Top Questions on Nernst Equation
Consider the cell $Pt _{( s )}\left| H _2( g , 1 atm )\right| H ^{+}( aq , 1 M )|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) Pt ( s )$ When the potential of the cell is $0.712 \, V$ at $298 \,K$, the ratio $\left[ Fe ^{2+}\right] /\left[ Fe ^{3+}\right]$ is ______(Nearest integer) Given : $Fe ^{3+}+ e ^{-}- Fe ^{2+}, E ^\theta Fe ^{3+}, Fe ^{2+} Pt =0771$ $\frac{2303 RT }{ F }=0.06 \,V$
- JEE Main - 2023
- Chemistry
- Nernst Equation
- The electrode potential of the following half cell at $298\, K$ $X \left| X ^{2+}(0.001 M ) \| Y ^{2+}(0.01 M )\right| Y$ is _____$\times 10^{-2} V$ (Nearest integer)
Given: $E _{ x ^{2 *} \mid x }^0=-2.36\, V$
$ E _{ Y ^{0-1 Y }}^0=+0.36\, V$
$\frac{2303 RT }{ F }=0.06 \,V$- JEE Main - 2023
- Chemistry
- Nernst Equation
- $P t(s)\left|II_2( g )(1 bar )\right| II^{+}(a q)(1 M)|| M^{3+}(a q), M^{+}(a q) \mid P t(s)$ The $E _{\text {cell }}$ for the given cell is $01115 V$ at $298 K$ when $\frac{\left[M^{+}(a q)\right]}{\left[M^{3+}(a q)\right]}=10^{\circ}$ The value of $a$ is _____ Given : $E ^\theta M ^{3+} M ^{+}=02 V$ $\frac{2303 R T}{F}=0059 V$
- JEE Main - 2023
- Chemistry
- Nernst Equation
- The logarithm of equilibrium constant for the reaction $Pd ^{2+}+4 Cl ^{-} \rightleftharpoons PdCl _4^{2-}$ is (Nearest integer) Given : $\frac{230RT }{ F }=006 V$ $Pd _{\text {(aq) }}^{2+}+2 e ^{-} \rightleftharpoons Pd ( s ) \quad E ^{\ominus}=083 V$ $PdCl _4^{2-} \text { (aq) }+2 e ^{-} \rightleftharpoons Pd ( s )+4 Cl ^{-} \text {(aq) } E ^{\ominus}=065 V$
- JEE Main - 2023
- Chemistry
- Nernst Equation
- Given $E^0_{Fe^{+3}/Fe^{+2}} = +0.76V$ and $E^0_{I_2/I^-} = +0.55V.$ The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is $\left[\frac{2.303RT}{F}=0.06\right]$
- KCET - 2020
- Chemistry
- Nernst Equation
View More Questions
Questions Asked in Manipal exam
- Pressure inside two soap bubbles are 1.01 atm and 1.03 atm. Ratio between their volumes is
- Manipal - 2013
- Surface tension
- Which of the following is the pribnow box?
- Manipal - 2013
- Transcription
- Who has written the book Genera Plantarum?
- Manipal - 2013
- Nomenclature and taxonomy
- There is loss in weight when mixture of $ L{{i}_{2}}C{{O}_{3}} $ and $ N{{a}_{2}}C{{O}_{3}}.10{{H}_{2}}O $ is heated strongly. This loss is due to
- Manipal - 2013
- anomalous behaviour of lithium
- The main function of clitellum is
- Manipal - 2013
- earthworm
View More Questions
Concepts Used:
Nernst Equation
This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.
