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A steady current of 1.5 amp flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of copper is $30 \times 10^{-5}$ g coulomb$^{-1}$, the mass of copper deposited on the electrode will be
A
0.50 g
B
0.67 g
C
0.27 g
D
0.40 g
Detailed Solution
By Faraday's first law of electrolysis: $m = ZIt$
Here $Z$ is the electrochemical equivalent of copper.
$Z = 30 \times 10^{-5}$ g C$^{-1}$, $I = 1.5$ A, $t = 10$ min $= 10 \times 60 = 600$ s
$m = 30 \times 10^{-5} \times 1.5 \times 10 \times 60$
$m = 30 \times 10^{-5} \times 900$
$m = 0.27$ g
Here $Z$ is the electrochemical equivalent of copper.
$Z = 30 \times 10^{-5}$ g C$^{-1}$, $I = 1.5$ A, $t = 10$ min $= 10 \times 60 = 600$ s
$m = 30 \times 10^{-5} \times 1.5 \times 10 \times 60$
$m = 30 \times 10^{-5} \times 900$
$m = 0.27$ g