If ${\rm{2}}\,\,{\rm{A}}$ of current is passed through ${\rm{C}}\,{\rm{u}}\,{\rm{S}}\,{{\rm{O}}_4}$ solution for ${\rm{32}}\,\,{\rm{s}}$, then the number of copper ions deposited at the cathode will be

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

\mathrm{2}\,\,\mathrm{A}

of current is passed through

\mathrm{C}\,\mathrm{u}\,\mathrm{S}\,{\mathrm{O}_4}

solution for

\mathrm{32}\,\,\mathrm{s}

, then the number of copper ions deposited at the cathode will be

View answer and explanation

Correct option: B

2\,\, \times \,\,{10^{20}}

Option B (

2 \times 10^{20}

) is correct because the number of copper ions is calculated using the formula q = it = n * 2e, where n is the number of ions.

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22.5\,cm