Three moles of electrons are passed through three solutions in succession containing ${A g N O}_{3}, {C u S O}_{4} a n d {A u C l}_{3}$ respectively. The molar ratio of amounts of cations reduced at cathode will be :

1 : 2; 3

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2 : 1 : 3

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3 : 2 : 1

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6 : 3 : 2

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Solution

The correct option is B $6 : 3 : 2$
${A g}^{+}, {C u}^{2 +}, {A u}^{3 +}$
$\frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$= 6 : 3 : 2$

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Similar questions

Q. Three mole of electrons are passed through three solutions in succession containing

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and

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respectively. The ratio of amounts of cations reduced at cathode will be___________.

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Three faraday of electricity is passed through three electrolytic cells connected in series containing $A g^{+}, C a^{2 +}$ and $A l^{3 +}$ ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is:

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Three moles of electrons are passed through three solutions in succession containing AgNO3,CuSO4andAuCl3 respectively. The molar ratio of amounts of cations reduced at cathode will be :

The correct option is B 6:3:2Ag+,Cu2+,Au3+11:12:13=6:3:2

Three moles of electrons are passed through three solutions in succession containing ${A g N O}_{3}, {C u S O}_{4} a n d {A u C l}_{3}$ respectively. The molar ratio of amounts of cations reduced at cathode will be :

The correct option is B $6 : 3 : 2$
${A g}^{+}, {C u}^{2 +}, {A u}^{3 +}$
$\frac{1}{1} : \frac{1}{2} : \frac{1}{3}$
$= 6 : 3 : 2$