Question

An electric current of 1.00 ampere is passed through an aqueous solution of Ni(NO3)2. How long will it take to plate out exactly 1.00 mol of nickel metal, assuming 100 percent current efficiency?

Answer 1

Answer: It takes 193000 seconds to plate out exactly 1.00 mol of nickel metal, assuming 100 percent current efficiency.

Explanation:

1 electron carry charge=$1.6\times 10^{-19}C$

1 mole of electrons contain= electrons

Thus  1 mole of electrons carry charge=$\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C=1 Faraday$

$Ni(NO_3)_2\rightarrow Ni^{2+}+2NO_3^-$

$Ni^{2+}+2e^-\rightarrow Ni$

$96500\times 2=193000Coloumb$ of electricity deposits 1.00 mole of nickel

$Q=I\times t$

where Q= quantity of electricity in coloumbs  = 193000 C

I = current in amperes  = 1.00 A

t= time in seconds = ?

$193000=1.00A\times t$

$t=193000s$

Thus it takes 193000 seconds to plate out exactly 1.00 mol of nickel metal, assuming 100 percent current efficiency.

Answer 2

Answer:

time required is 193000 s

Explanation:

$Ni^{2+}(aq)+2e^- \rightarrow Ni(s)$

From the above it is clear that for plating out 1 atom of Ni, 2 electrons are needed.

Therefore, for plating out 1 mol of nickel metal, 2 mols of electrons are needed.

Charge of 1 mol of electron = 96500 C = 1 Faraday

Therefore, charge of 2 moles of electron = 2 × 96500

                                                                     = 193,000‬ C

Relation between current, charge and time is as follows:

Q = I × t

193000 = 1.00 × t

t = 193000 s