A) 0.1
B) 2.0
C) 1.0
D) 0.5
Correct Answer: C
Solution :
Reduction at cathode:\[2{{H}^{+}}(aq)+2{{e}^{-}}\to {{H}_{2}}(g)\] At N.T.P, 22.4 L (or 22400 mL) of \[{{H}_{2}}=1\] mole of \[{{H}_{2}}\] 112 mL of \[{{H}_{2}}=\frac{112}{22400}\times 1=0.005\] mole of \[{{H}_{2}}\] Moles of \[{{H}_{2}}\]produced \[=\frac{I(A)\times t(s)}{96500(C/mol{{e}^{-}})}\times \] mole ratio 0.005mol\[=\frac{I(A)\times 965s}{96500(C/mol{{e}^{-}})}\times \frac{1\,\,mol\,\,{{H}_{2}}}{2\,\,mol\,\,{{e}^{-}}}\] \[I=1A\]
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