Paragraph: Tollen's reagent is used for the detection of aldehyde when a solution of AgNO_3 is added to glucose with NH_4 OH then gluconic acid is formed array rr Ag^ + +e^ - arrow Ag & E_ red ^ =0.8 ~V C_6 H_ 12

Correct answer: C. E_ red will increase by a factor of 0.65 from E_ red ^ On increasing concentration of NH_3 , the concentration of H^ + ion decreases, therefore aligned E_ red & =E_ red ^ ^ -(0.0592)/(n) _ 10 [H^ + ]^2 & =0-(0.0592)/(2) ×2 _ 10 10^ -11 ( pH=11 ; [H^ + ]=10^ -11 M) & =-0.0592 ×11=0.6512 V=0.65 V aligned E_ red increased by a factor of 0.65 from E_ red ^ ^ .

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Chemistry
Chapter: Electrochemistry

JEE Advanced 2006 · Chemistry · Electrochemistry

Paragraph: Tollen's reagent is used for the detection of aldehyde...

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

\begin{array}{rr}\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}\end{array}

[Use

2.303 \times \frac{R T}{F}=0.0592

and

\frac{F}{R T}=38.92

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?

Options

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}

Correct answer

C. $E_{\text{red }}$ will increase by a factor of $0.65$ from $E_{$ $\text{red }^{$ \circ}

Step-by-step solution

On increasing concentration of

\mathrm{NH}_3

, the concentration of

\mathrm{H}^{+}

ion decreases, therefore

\begin{aligned} E_{\mathrm{red}} & =E_{\mathrm{red}}^{{}^{\circ}}-\frac{0.0592}{n} \log _{10}\left[\mathrm{H}^{+}\right]^2 \\ & =0-\frac{0.0592}{2} \times2 \log _{10} 10^{-11} \quad\left(\because \mathrm{pH}=11 ; \therefore\left[\mathrm{H}^{+}\right]=10^{-11} \mathrm{M}\right)\\ & =-0.0592 \times11=0.6512 \mathrm{ V}=0.65 \mathrm{ V} \end{aligned}

E_{\text{red }}

increased by a factor of

0.65

from

E_{\text{red }}^{{}^{\circ}}$.

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