Question
The standard emf of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction would be (Given $\left.F=96500 \mathrm{C} \mathrm{mol}^{-1} ; \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$(a) $2.0 \times 10^{11}$(b) $4.0 \times 10^{12}$(c) $1.0 \times 10^{2}$(d) $1.0 \times 10^{10}$
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0591}{n} \log Q\] where \(E\) is the cell potential, \(E^0\) is the standard cell potential, \(n\) is the number of electrons transferred in the cell reaction, and \(Q\) is the reaction quotient. Show more…
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The standard e.m.f. of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction would be (Given $\left.\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1} ; \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$ a. $2.0 \times 10^{11}$ b. $4.0 \times 10^{12}$ c. $1.0 \times 10^{2}$ d. $1.0 \times 10^{10}$
For a cell reaction involving two electrons, the standard $\mathrm{emf}$ of the cell is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction at $25^{\circ} \mathrm{C}$ will be: (a) $1 \times 10^{-10}$ (b) $29.5 \times 10^{-2}$ (c) 10 (d) $1 \times 10^{10}$
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}\right.$ $\left.=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$ (a) $1.0 \times 10^{30}$ (b) $1.0 \times 10^{1}$ (c) $1.0 \times 10^{5}$ (d) $1.0 \times 10^{10}$
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