Question

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

• A. $E-E^{\circ} = +\frac{2RT}{nF}$
• B. $E-E^{\circ} = -\frac{2RT}{nF}$
• C. $E-E^{\circ} = \frac{RT}{nF}$
• D. $E-E^{\circ} = \frac{-RT}{nF}$

Answer 1

Answer: (A)

Oxidised state $\left(O\right) + ne^{-} \rightleftharpoons $ Reduced state (R) From Nernst eqn.
$E-E^{\circ} -\frac{RT}{nF}$ In $\frac{\left[R\right]}{\left[O\right]}$
$E -E^{\circ} =\frac{RT}{-nF}$ In $\frac{1}{e^{2}} = \frac{RT}{-nF}$ In $\left(e^{-2}\right)$
$E-E^{\circ} =- \frac{\left(-2\right)RT}{nF}\, lne$
$E -E^{\circ} = + \frac{RT}{nF} \left(\because lne = 1\right)$