Unit 3 Electrochemistry (Intext Questions-4)
Intext Questions
3.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
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Answer
$I=0.5 \mathrm{~A}$
$t=2$ hours $=2 \times 60 \times 60 \mathrm{~s}=7200 \mathrm{~s}$
Thus, $Q=I t$
$=0.5 \mathrm{~A} \times 7200 \mathrm{~s}$
$=3600 \mathrm{C}$
We know that $96487 \mathrm{C}=6.023 \times 10^{23}$ number of electrons.
Then,
$3600 \mathrm{C}=\frac{6.023 \times 10^{23} \times 3600}{96487}$ number of electrons
$=2.25 \times 10^{22}$ number of electrons
Hence, $2.25 \times 10^{22}$ number of electrons will flow through the wire.
3.11 Suggest a list of metals that are extracted electrolytically.
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Answer
Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically.
3.12 What is the quantity of electricity in coulombs needed to reduce $1 \mathrm{~mol}$ of $\mathrm{Cr_2} \mathrm{O_7}^{2-} $ Consider the reaction: $\mathrm{Cr_2} \mathrm{O_7}{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H_2} \mathrm{O}$
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Answer
The given reaction is as follows:
$\mathrm{Cr_2} \mathrm{O_7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H_2} \mathrm{O}$
Therefore, to reduce 1 mole of $\mathrm{Cr_2} \mathrm{O_7}^{2-}$, the required quantity of electricity will be:
$=6 \mathrm{~F}$
$=6 \times 96487 \mathrm{C}$
$=578922 \mathrm{C}$
