Redox Reactions And Electrochemistry Question 466
Question: A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of $ 0.250mol $ of metal X at the cathode. The oxidation state of X in the molten salt is: $ ( F=96,500C ) $
Options:
A) $ {1^{+}} $
B) $ {2^{+}} $
C) $ {3^{+}} $
D) $ {4^{+}} $
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Answer:
Correct Answer: C
Solution:
[c] According to Faraday’s first law of electrolysis $ W=\frac{E\times i\times t}{96500} $ Where E = equivalent weight $ =\frac{mol.mass,of,metal(M)}{oxidationstateofmetal( x )} $
Substituting the value in the formula $ W=\frac{M}{x}\times \frac{i\times t}{96500} $ or $ x=\frac{M}{W}\times \frac{i\times t}{96500}=\frac{10\times 2\times 60\times 60}{96500\times 0.250}=3 $ $ [ Given:no.ofmoles=\frac{M}{W}=0.250 ] $ Hence oxidation state of metal is (+3)
BETA
