The equilibrium constant for the redox change -2AgNH32- - C6H12 O6 - H2O - C6H12 O7 - 2Ags - C6H12 O2 - 2H- is-

The correct option is D 5.15 × 10^9 [2Ag(NH_3)_2^+ + C_6H_12 O_6 + H_2O nbsp;→ C_6H_12 O_7 + 2Ag(s) + C_6H_12 O_2 + 2H^+] E = E^o_OP_Glucose ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

The equilibri...

Question

The equilibrium constant for the redox change
$[2 A g (N H_{3})_{2}^{+} + C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 A g (s) + C_{6} H_{12} O_{2} + 2 H^{+}]$ is:

5.15 \times 10^{9}

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4, 2 \times 10^{5}

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1.72 \times 10^{- 11}

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4.2 \times 10^{- 5}

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Solution

The correct option is D $5.15 \times 10^{9}$
$[2 A g (N H_{3})_{2}^{+} + C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 A g (s) + C_{6} H_{12} O_{2} + 2 H^{+}]$
$E = E_{O P_{G l u c o s e}}^{o} + E_{R P_{A g (N H_{3})_{2}^{+} / A g}}^{o} = - 0.05 + 0.337 = 0.287 V$
According to Nernst equation:
$E = E^{o} - 0.0591 l o g Q / n$
And at eqm. $E = 0$
so $E^{o} = \frac{2.303 R T}{n F} l o g K = \frac{0.0591}{2} l o g K$
$∴ l o g K = \frac{2 \times 0.287}{0.0591} = 9.71$
$K = 5.15 \times 10^{9}$

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Q. For

2 A g^{+} + C_{6} H_{12} O_{6} + H_{2} O \to 2 A g (s) + C_{6} H_{12} O_{7} + 2 H^{+}

, ln K of the change is:

2 A g^{+} + C_{6} H_{12} O_{6} + H_{2} O \to 2 A g (s) + C_{6} H_{12} O_{7} + 2 H^{+}

Find

ln K

this reaction.

Q. Tollen's reagent is used for the detection of aldehydes. When a solution of

A g N O_{3}

is added to glucose with

N H_{4} O H

, then gluconic acid is formed.

A g^{+} + e^{-} \to A g; E_{r e d}^{o} = 0.80 V

C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 H^{+} + 2 e^{-}; E_{o x i}^{o} = - 0.05 V

A g (N H_{3})_{2}^{+} + e^{-} \to A g (s) + 2 N H_{3}; E_{r e d}^{o} = 0.337 V

Find

log K

of the reaction?

2 A g^{+} + C_{6} H_{12} O_{6} + H_{2} O \to 2 A g (s) + C_{6} H_{12} O_{7} + 2 H^{+}

(Take : 2.303 \times \frac{R T}{F} = 0.06 at 298 K)

Q. Tollen's reagent is used for the detection of aldehydes. When a solution of

A g N O_{3}

is added to glucose with

N H_{4} O H

, then gluconic acid is formed.

A g^{+} + e^{-} \to A g; E_{r e d}^{o} = 0.80 V

C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 H^{+} + 2 e^{-}; E_{o x i}^{o} = - 0.05 V

A g (N H_{3})_{2}^{+} + e^{-} \to A g (s) + 2 N H_{3}; E_{r e d}^{o} = 0.337 V

Find

log K

of the reaction?

2 A g^{+} + C_{6} H_{12} O_{6} + H_{2} O \to 2 A g (s) + C_{6} H_{12} O_{7} + 2 H^{+}

(Take : 2.303 \times \frac{R T}{F} = 0.06 at 298 K)

Tollen’s reagent is used for the detection of aldehyde, when a solution of $A g N O_{3}$ is added to glucose with $N H_{4} O H$ then gluconic acid is formed.

$A g^{+} + e^{-} \to A g;$ $E_{r e d}^{\circ} = 0.8 V$

$C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7}$ (Gluconic acid) $+ 2 H^{+} + 2 e; E_{o x d}^{\circ} = - 0.05 V$

$A g (N H_{3})_{2}^{+} + e^{-} \to A g (S) + 2 N H_{3}; E^{\circ} = - 0.337 V$

[Use $2.303 \times \frac{R T}{F} = 0.0592$ and $\frac{F}{R T} = 38.92$ at 298 K]

(i) $2 A g^{+} + C_{6} H_{12} O_{6} + H_{2} O \to 2 A g (s) + C_{6} H_{12} O_{7} + 2 H^{+}$ . Find ln K of this reaction.

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The equilibrium constant for the redox change[2Ag(NH3)+2+C6H12O6+H2O→C6H12O7+2Ag(s)+C6H12O2+2H+] is:

The correct option is D 5.15×109[2Ag(NH3)+2+C6H12O6+H2O→C6H12O7+2Ag(s)+C6H12O2+2H+]E=EoOPGlucose+EoRPAg(NH3)+2/Ag=−0.05+0.337=0.287VAccording to Nernst equation:E=Eo−0.0591logQ/nAnd at eqm. E=0so Eo=2.303RTnFlogK=0.05912logK∴logK=2×0.2870.0591=9.71K=5.15×109

The equilibrium constant for the redox change
$[2 A g (N H_{3})_{2}^{+} + C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 A g (s) + C_{6} H_{12} O_{2} + 2 H^{+}]$ is: