A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At. wt. = 177). The oxidation state of the metal in the metal salt is:

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Solution

The correct option is C 3
$\frac{Q}{F} = m o l e s \times n_{f} \frac{i \times t}{F} = m o l e s \times n_{f} \frac{2 \times 5 \times 3600}{96500} = \frac{22.2}{177} n_{f} \Rightarrow n_{f} = 3$
Thus, oxidation state of metal in the metal salt is +3.

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Similar questions

A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At. wt. = 177). The oxidation state of the metal in the metal salt is:

Q. A current of

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A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At. wt. = 177). The oxidation state of the metal in the metal salt is:

The correct option is C 3QF=moles×nfi×tF=moles×nf2×5×360096500=22.2177nf⇒nf=3 Thus, oxidation state of metal in the metal salt is +3.

The correct option is C 3
$\frac{Q}{F} = m o l e s \times n_{f} \frac{i \times t}{F} = m o l e s \times n_{f} \frac{2 \times 5 \times 3600}{96500} = \frac{22.2}{177} n_{f} \Rightarrow n_{f} = 3$
Thus, oxidation state of metal in the metal salt is +3.