Question

Mass of Nickel deposited at cathode by passing 9.65 ampere current for a period of 100 minutes through molten Nickel is (Atomic mass of Ni = 58.7u)
A. 35.22 g
B. 17.61 g
C. 8.81 g
D. 22.51g

Answer 1

Write the half cell reaction for the deposition nickel. Using the current and time calculate the charge. Using the charge, calculate the moles of electrons. Using the moles of electrons and moles electrons transfer in half cell reaction calculate the moles of nickel. Finally, convert the moles of nickel to the mass of nickel.

Formula Used: $Q = I \times t$

Complete answer:
The half cell reaction for the deposition nickel is as follows:
${\text{N}}{{\text{i}}^{{\text{2 + }}}}{\text{(aq) + 2}}{{\text{e}}^{\text{ - }}} \to {\text{ Ni (s)}}$
Now, using the given value of current and time we can calculate the charge as follows:
$Q = I \times t$
where,
$Q$ = Charge in coulomb
$I$ = Current in ampere = 9.65 ampere
$t$ = time in seconds = 100 minutes = 6000 s (As 1 min = 60 s)
Now, substitute 9.65 amperes for current, 6000s for time and calculate the charge.
$Q = I \times t$
$\Rightarrow Q = \,9.65{\text{ ampere}} \times {\text{ 6000 s = 57900 C}}$
Now, using the relation between charge and moles of electrons we can calculate the moles of electron transfer as follows:
$\text{96500 C = 1 mol electrons}$
$\Rightarrow {\text{ 57900 C}} \times \dfrac{{1{\text{ mol electron}}}}{{96500{\text{ C}}}} = 0.6{\text{ mol electron }}$
From the half cell reaction, we can say that 1 mole of nickel is deposited when 2 moles of electrons are transferred. So using the calculated moles of electrons we can calculate the moles of nickel deposited as follows:
$0.6{\text{ mol electron }} \times \dfrac{{{\text{1 mole of Ni}}}}{{2{\text{ mol of }}}} = 0.3{\text{ mol Ni}}$
Now, using calculated moles of nickel and its given atomic mass we can calculate the mass of nickel deposited as follows:
${\text{Mass of Ni = mol of Ni }} \times {\text{ Atomic mass of Ni}}$
Substitute 0.3 mol for mol of nickel, 58.7u for the atomic mass of nickel and calculate the mass of nickel.
$\Rightarrow {\text{Mass of Ni = }}0.3{\text{ mol }} \times {\text{ 58}}{\text{.7 u = 17}}{\text{.61 g }}$
Thus, the mass of nickel deposited is $\text{17.61g}$.

**Hence,the correct option is (B) 17.61g

Note:**
For calculating charge using current and time always use the current in ampere and time in seconds. Also, it is important to write the half cell reaction correctly as moles of metal deposited depend on moles of electrons transferred.