If the ratio of composition of oxidised and reduced species in electrochemical cell is given as ([O]/[R]) =e2 the correct potential difference will be

Question

If the ratio of composition of oxidised and reduced species in electrochemical cell is given as ([O]/[R]) =e2 the correct potential difference will be (A) E-E° = +(2RT/nF) (B) E-E° = -(2RT/nF) (C) E-E° = (RT/nF) (D) E-E° = (-RT/nF)

Tardigrade Admin · Accepted Answer

Oxidised state (O) + ne- leftharpoons Reduced state (R) From Nernst eqn. E-E° -(RT/nF) In ([R]/[O]) E -E° =(RT/-nF) In (1/e2) = (RT/-nF) In (e-2) E-E° =- ((-2)RT/nF) lne E -E° = + (RT/nF) (∵ lne = 1)

Q. If the ratio of composition of oxidised and reduced species in electrochemical cell is given as $\frac{[ O ]}{[ R ]} = e^{2}$ the correct potential difference will be

$E - E^{\circ} = + \frac{2 RT}{n F}$

$E - E^{\circ} = - \frac{2 RT}{n F}$

$E - E^{\circ} = \frac{RT}{n F}$

$E - E^{\circ} = \frac{- RT}{n F}$

Q. If the ratio of composition of oxidised and reduced species in electrochemical cell is given as [R][O]​=e2 the correct potential difference will be

E−E∘=+nF2RT​

E−E∘=−nF2RT​

E−E∘=nFRT​

E−E∘=nF−RT​

Oxidised state (O)+ne−⇌ Reduced state (R) From Nernst eqn. E−E∘−nFRT​ In [O][R]​ E−E∘=−nFRT​ In e21​=−nFRT​ In (e−2) E−E∘=−nF(−2)RT​lne E−E∘=+nFRT​(∵lne=1)

Q. If the ratio of composition of oxidised and reduced species in electrochemical cell is given as $\frac{[ O ]}{[ R ]} = e^{2}$ the correct potential difference will be

$E - E^{\circ} = + \frac{2 RT}{n F}$

$E - E^{\circ} = - \frac{2 RT}{n F}$

$E - E^{\circ} = \frac{RT}{n F}$

$E - E^{\circ} = \frac{- RT}{n F}$