A current of 10.0 A flows for 2.00 hrs through an electrolytic cell containing a molten salt of metal $X$ . This results in the decomposition of 0.250 mol of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is: $(F = 96500 C)$

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Solution

The correct option is B 3
$t = 2 \times 60 \times 60 = 7200 s$
Charge $=$ current time $= 10 A \times 7200 s = 72000 C$
According to the reaction: $X^{n +} (a q .) + n e^{-} \to X (s)$
We require $n F$ or $n 96500 C$ to deposit 1 mol of $X$ .
For 0.25 mole of deposition we require $n \times 24125$ $C$ .
Hence, $n = \frac{72000}{24125} = 3$

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A current of 10.0 A flows for 2.00 hrs through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is:(F=96500C)

The correct option is B 3t=2×60×60=7200sCharge = current time =10 A×7200 s=72000 CAccording to the reaction: Xn+(aq.)+ne−→X(s)We require nF or n96500 C to deposit 1 mol of X.For 0.25 mole of deposition we require n×24125 C.Hence, n=7200024125=3

A current of 10.0 A flows for 2.00 hrs through an electrolytic cell containing a molten salt of metal $X$ . This results in the decomposition of 0.250 mol of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is: $(F = 96500 C)$