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Physics-

General

Easy

Question

If 1 A of current is passed through $C u S O subscript 4 end subscript$ solution for 10 seconds, then the number of copper ions deposited at the cathode will be about

$1.6 \times 1 0^{19}$
$3.1 \times 1 0^{19}$
$4.8 \times 1 0^{19}$
$6.2 \times 1 0^{19}$

The correct answer is: $3.1 cross times 1 0 to the power of 19 end exponent$

Total charge supplied $equals 1 cross times 10 equals 10 C$
$because$ 2 electronic charge $left parenthesis 3.2 cross times 1 0 to the power of negative 19 end exponent C right parenthesis$ liberates one $C u to the power of plus plus end exponent$ ion
\ Number of $C u to the power of plus plus end exponent$ ions liberated by 10 C charge
$equals fraction numerator 1 over denominator 3.2 cross times 1 0 to the power of negative 19 end exponent end fraction cross times 10 equals 3.1 cross times 1 0 to the power of 19 end exponent$

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