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Standard XII

Chemistry

Potassium Permanganate

A constant cu...

Question

A constant current was passed through a solution of $A u C l_{4}^{⊖}$ ion between gold electrodes. After a period of $10.0 m i n$ , the increase in the weight of cathode was $1.314 g$ . The total charge passed through solution is: (atomic weight of $A u C l_{4}^{⊖} = 339$ )

1.16, \times 10^{- 2} F

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3.5 \times 10^{- 2} F

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2 \times 10^{- 2} F

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4 \times 10^{- 3} F

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Solution

The correct option is C $2 \times 10^{- 2} F$
$A u^{3 +} + 3 e^{-} \to A u$
$\frac{W}{197 / 3} = \frac{C h a r g e}{96500}$
$∴ C h a r g e = \frac{1.314 \times 3 \times F}{197} = 0.02 F$
= $2 \times 10^{- 2} F$

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A constant current was passed through a solution of AuCl⊖4 ion between gold electrodes. After a period of 10.0min, the increase in the weight of cathode was 1.314g. The total charge passed through solution is: (atomic weight of AuCl⊖4=339)

The correct option is C 2×10−2FAu3++3e−→AuW197/3=Charge96500 ∴Charge=1.314×3×F197=0.02F = 2×10−2F

A constant current was passed through a solution of $A u C l_{4}^{⊖}$ ion between gold electrodes. After a period of $10.0 m i n$ , the increase in the weight of cathode was $1.314 g$ . The total charge passed through solution is: (atomic weight of $A u C l_{4}^{⊖} = 339$ )

The correct option is C $2 \times 10^{- 2} F$
$A u^{3 +} + 3 e^{-} \to A u$
$\frac{W}{197 / 3} = \frac{C h a r g e}{96500}$
$∴ C h a r g e = \frac{1.314 \times 3 \times F}{197} = 0.02 F$
= $2 \times 10^{- 2} F$