JEE Advanced · Chemistry · 17. Electrochemistry
Paragraph:
Tollen's reagent is used for the detection of aldehyde when a solution of \(\mathrm{AgNO}_3\) is added to glucose with \(\mathrm{NH}_4 \mathrm{OH}\) then gluconic acid is formed
\(\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \)
\( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)\) \(+~2 \mathrm{H}^{+}+2 e^{-} ; -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \)
\( \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ;\) \(E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\)
[Use \(2.303 \times \frac{R T}{F}=0.0592\) and \(\frac{F}{R T}=38.92\) at \(298 \mathrm{~K}\) ]
Question:
When ammonia is added to the solution, \(\mathrm{pH}\) is raised to 11 . Which half-cell reaction is affected by \(\mathrm{pH}\) and by how much?
- A \(E_{\text {oxd }}\) will increase by a factor of \(0.65\) from \(E_{\text {oxd }}^{\circ}\)
- B \(E_{\text {oxd }}\) will decrease by a factor of \(0.65\) from \(E_{\text {oxd }}^{\circ}\)
- C \(E_{\text {red }}\) will increase by a factor of \(0.65\) from \(E_{\text {red }}^{\circ}\)
- D \(E_{\text {red }}\) will decrease by a factor of \(0.65\) from \(E_{\text {red }}^{\circ}\)
Answer & Solution
Correct Answer
(C) \(E_{\text {red }}\) will increase by a factor of \(0.65\) from \(E_{\text {red }}^{\circ}\)
Step-by-step Solution
Detailed explanation
On increasing concentration of \(\mathrm{NH}_3\), the concentration of \(\mathrm{H}^{+}\)ion decreases, therefore
\(E_{\mathrm{red}} =E_{\mathrm{red}}^{\circ}-\frac{0.0592}{n} \log _{10}\left[\mathrm{H}^{+}\right]^2 \)
\( =0-\frac{0.0592}{2} \times 2 \log _{10} 10^{-11} \quad\)\(\left(\because \mathrm{pH}=11 ; \therefore\left[\mathrm{H}^{+}\right]=10^{-11} \mathrm{M}\right) \)
\( =-0.0592 \times 11=0.6512 \mathrm{~V}=0.65 \mathrm{~V}\)
\(E_{\text {red }}\) increased by a factor of \(0.65\) from \(E_{\text {red }}^{\circ}\).
\(E_{\mathrm{red}} =E_{\mathrm{red}}^{\circ}-\frac{0.0592}{n} \log _{10}\left[\mathrm{H}^{+}\right]^2 \)
\( =0-\frac{0.0592}{2} \times 2 \log _{10} 10^{-11} \quad\)\(\left(\because \mathrm{pH}=11 ; \therefore\left[\mathrm{H}^{+}\right]=10^{-11} \mathrm{M}\right) \)
\( =-0.0592 \times 11=0.6512 \mathrm{~V}=0.65 \mathrm{~V}\)
\(E_{\text {red }}\) increased by a factor of \(0.65\) from \(E_{\text {red }}^{\circ}\).
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