Determing Valence Electrons

How many seconds does it take to deposit 0.94 g of Ni on a decorative drawer handle when 14.9 A is passed through a Ni(NO3)2 sol

Goshia [24]

Answer:

Time take to deposit Ni is 259.02 sec.

Explanation:

Given:

Current $I = 14.9$ A

Faraday constant $= 96485$ $\frac{C}{mole}$

Molar mass of Ni $= 58.69$ $\frac{g}{mole}$

Mass of Ni $= 0.94$ g

First find the no. moles in Ni solution,

Moles of Ni $= \frac{0.94}{58.69}$

$= 0.02$ mol

From the below reaction,

$Ni^{2+} + 2e$ ⇆ $Ni_{(s)}$

Above reaction shows "1 mol of $Ni^{2+}$ requires 2 mol of electron to form 1 mol of $Ni_{(s)}$ "

So for finding charge flow in this reaction we write,

$=$ $0.02 \times \frac{2 }{1 } \times 96485 \frac{C}{mol}$

Charge flow $= 3859.4$ C

For finding time of reaction,

$I = \frac{q}{t}$

Where $q =$ charge flow

$t = \frac{q}{I}$

$t = \frac{3859.4}{14.9}$

$t = 259.02$ sec

Therefore, time take to deposit Ni is 259.02 sec.

3 0

2 years ago

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Lera25 [3.4K]

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3 years ago

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2 years ago

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Hope this helps!

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