Electrodeposition of copper: what mass of copper can be deposited in 1.00 hr by a current of 1.62 A?

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Electrodeposition of copper: what mass of copper can be deposited in 1.00 hr by a current of 1.62 A?

$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$ $Cu^(2+)(aq)+2e^(-)\rarrCu(s)$

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Answer 1 · 2018-06-27T03:03:34

This can be done with a unit conversion.

Each ${Cu}^{2 +}$ $"Cu"^(2+)$ receives $2$ $2$ electrons in this half-reaction, so $n = {2 mol e}^{-} /mol atom$ $n = "2 mol e"^(-)"/mol atom"$ .
$1.62 A$ $"1.62 A"$ $=$ $=$ $1.62 C/s$ $"1.62 C/s"$ is the current.
$F = {96485 C/mol e}^{-}$ $F = "96485 C/mol e"^(-)$ is the Faraday constant.

So, in $1.00 hr$ $"1.00 hr"$ ,

$1.00 cancel"hr" xx (60 cancel"min")/(cancel"1 hr") xx (60 cancel"s")/cancel"1 min" = "3600 s"$

has passed, and

$3600 cancel"s" xx (1.62 cancel"C")/(cancel"s") xx (cancel("1 mol e"^(-)))/(96485 cancel"C") xx (cancel"1 mol Cu")/(cancel("2 mol e"^(-))) xx ("63.55 g Cu")/(cancel"1 mol Cu")$

$=$ $color(blue)("1.92 g Cu"(s))$

can be deposited into the cathode.

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Electrodeposition of copper: what mass of copper can be deposited in 1.00 hr by a current of 1.62 A?

$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$ $Cu^(2+)(aq)+2e^(-)\rarrCu(s)$

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Electrodeposition of copper: what mass of copper can be deposited in 1.00 hr by a current of 1.62 A?

Cu^(2+)(aq)+2e^(-)\rarrCu(s)Cu2+(aq)+2e−→Cu(s)Cu^(2+)(aq)+2e^(-)\rarrCu(s)

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$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$ $Cu^(2+)(aq)+2e^(-)\rarrCu(s)$