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\mathrm{Ag}_2 \mathrm{~S}(s)+2 e^{-} \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq})
The above half cell reaction was buffered at \mathrm{pH} value of \mathrm{3}. The same is saturated with \mathrm{0.01 \mathrm{M}\ \mathrm{H}_2 \mathrm{S}\ (aq)}. What is the potential of the reaction if \mathrm{k_{s p\left(A_{g_2} S\right)}=10^{-49}} and \mathrm{k_{a_1} \cdot K_{a_2}=10^{-22}}.

Option: 1

\mathrm{-0.13\ V}


Option: 2

\mathrm{1.3\ V}


Option: 3

\mathrm{2.89\ V}


Option: 4

\mathrm{-0.013\ V}


Answers (1)

best_answer

\mathrm{\begin{aligned} \mathrm{H}_2 \mathrm{~S} & \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-} \\ \mathrm{K}_{a_1,} \cdot \mathrm{K}_{a_2} & =\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]} \end{aligned}}
\mathrm{\begin{aligned} 10^{-22} \times 0.01 & =\left(10^{-3}\right)^2\left[\mathrm{~S}^{2-}\right] \\ {\left[\mathrm{S}^{2-}\right] } & =10^{-18} \end{aligned}}
\mathrm{E_{S^{2-}\left|\mathrm{Ag}_2 S\right| \mathrm{Ag}}=E_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0=\frac{0.06}{2} \log \frac{\left[\mathrm{S}^{-2}\right]}{\mathrm{K}_{\text {sp }}}}
\mathrm{\begin{aligned} & =0.80-\frac{0.06}{2} \log \frac{10^{-18}}{10^{-49}} \\ & =0.80-0.03 \log 10^{31} \\ & =0.80-0.93 \\ & =-0.13 \mathrm{~V} \end{aligned}}

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Deependra Verma

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