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Question

Tollen's reagent is used for the detection of aldehyde when a solution of $$Ag{NO}_{3}$$ is added to glucose with $${NH}_{4}OH$$ then fluconic acid is formced
$${Ag}^{+}+{e}^{-}\rightarrow Ag$$;
$${E}_{red}^{o}=0.8V$$
$${C}_{6}{H}_{12}{O}_{6}+{H}_{2}O\rightarrow {C}_{6}{H}_{12}{O}_{7}$$ (Gluconic acid)$$+2{H}^{+}+2{e}^{-}$$; $${E}_{red}^{o}=-0.05V$$
$$Ag{({NH}_{3})}_{2}+{e}^{-}\rightarrow Ag(s)+2{NH}_{3}$$;
(Use $$2.202\times \cfrac{RT}{F}=0.0592$$ and $$\cfrac{F}{RT}=38.92 $$at $$298K$$)
$$2{Ag}^{+}+{C}_{6}{H}_{22}{O}_{6}+{H}_{2}O\rightarrow 2Ag(s)+{C}_{6}{H}_{212}{O}_{7}+2{H}^{+}$$
Find $$\ln {K}$$ of this reaction?


A
66.13
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B
58.38
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C
28.30
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D
46.29
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Solution

The correct option is B $$58.38$$
Here the answer is wrong, and the correct answer is option(B).

Solution:

$${ E }_{ cell }^{ 0 }=\frac { RT }{ nf } \quad ln\quad K$$

$$(0.8\quad -\quad 0.05)\quad =\frac { 1 }{ 2 } X\quad \frac { 0.0592 }{ 2.303 } ln\ K$$


$$ln\quad K=\frac { (0.8-0.05)X\quad 2X2.303 }{ 0.0592 } =58.38$$

Hence,58.38 is the correct answer.


Chemistry

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