Electrochemistry Result Question 33
Passage
Tollen’s reagent is used for the detection of aldehydes. When a solution of $\mathrm{AgNO}_3$ is added to glucose with $\mathrm{NH}_4 \mathrm{OH}$, then silver mirror is formed.
$ \begin{alignedat} & \mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; \quad E_{\text {red }}^{\circ}=0.80 \mathrm{V} \\ & \mathrm{C}6 \mathrm{H}{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow {\mathrm{C}6 \mathrm{H}{12} \mathrm{O}7}{\text { (Gluconic acid) }} +2 \mathrm{H}^{+}+2 e^{-} \text {; } \quad E{\text {oxi }}^{\circ}=-0.05 \mathrm{V} \\ & \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \longrightarrow \mathrm{Ag}(s)+2 \mathrm{NH}3 ; \quad E{\text {red }}^{\circ}=0.337 \mathrm{V} \\ \end{aligned} $
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $\frac{F}{R T}=38.92$ at 298 K ]
$ (2006,3 \times 4 M=12 \mathrm{M}) $
33. When ammonia is added to the solution, pH is raised to 11 . Which half-cell reaction is affected by pH and by how much?
(a) $E_{\text {oxi }}$ will increase by a factor of $0.65$ from $E_{\text {oxi }}^{\circ}$
(b) $E_{\text {oxi }}$ will decrease by a factor of $ 0.65$ from $E_{\text {oxi }}^{\circ}$
(c) $E_{\text {red }}$ will increase by a factor of $0.65$ from $E_{\text {red }}^{\circ}$
(d) $E_{\mathrm{red}}$ will decrease by a factor of $0.65$ from $E_{\text{red}}^{\mathrm{red}}$
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Answer:
Correct Answer: 33. ( c )
Solution:
- On increasing concentration of $\mathrm{NH}_3$, the concentration of $\mathrm{H}^{+}$ion decreases, therefore,
$ \begin{alignedat} E_{\text {red }} & =E_{\text {red }}^{\circ}-\frac{0.0592}{2} \log \left[\mathrm{H}^{+}\right]^2=0-\frac{0.0592}{2} \times 2 \log 10^{-11} \\ & =0.65 \mathrm{V} \end{aligned} $
i.e. $E_{\text {red }}$ increases by $0.65$ $V$ .
BETA
