Chemistry - Electrochemistry Question with Solution | TestHub

Question Explanation:

Calculate the change in cell EMF when the concentration of iodide ions is doubled.

Concept:

Nernst Equation and Concentration effect on EMF.

Solution:

Half-reaction: ${\mathrm{I}}_2+2{\mathrm{e}}^{-}\to 2{\mathrm{I}}^{-}$

Nernst Equation (Cathode): ${\mathrm{E}}_{\text{red }}={\mathrm{E}}_{\text{red }}^0-\frac{0.0591}{2}\log {\left[{\mathrm{I}}^{-}\right]}^2$

Change in EMF ( $\Delta \mathrm{E}$ ): When $\left[{\mathrm{I}}^{-}\right]$is doubled ( $2\times 0.1\mathrm{M}$ ): $\Delta \mathrm{E}=-\frac{0.0591}{2}\log {\left(\frac{{\left[{\mathrm{I}}^{-}\right]}_{\text{final }}}{{\left[{\mathrm{I}}^{-}\right]}_{\text{initial }}}\right)}^2$ $\Delta \mathrm{E}=-0.0591\log (2)\approx -0.0591\times 0.3010\approx -0.0178\mathrm{V}$

New EMF: $0.535\mathrm{V}-0.018\mathrm{V}=0.517\mathrm{V}$

Final Answer: Option (A)