Question

How many moles of Pt will be deposited on cathode when 0.80 F electricity is passed through 1M solution of $P{t^{4 + }}$.
(A) 0.80 mol
(B) 8 mol
(C) 0.20 mol
(D) 1 mol

Answer 1

To solve the above question we can use the Faraday’s first law of electrolysis, we can write equation as:
$n = \dfrac{Q}{{{n_f} \times F}}$
Here, n is number of moles, Q is total charge flowing in circuit, ${n_f}$ is n-factor or number of electrons flowing and F is Faraday constant.

Complete step by step solution:
We know from Faraday’s First Law of Electrolysis that the amount of substance deposited on electrodes is directly proportional to charge flowing in circuit.
$m\propto Q$
Here, $m$ is the mass of the substance deposited and $Q$ is the charge flowing in the circuit.
We make proportionality sign into equal by putting $Z$, which is electrochemical equivalent.
$m = Z.Q$ -------- (1)
Now, Z which is electrochemical equivalent can be written as:
$Z = \dfrac{M}{{{n_f} \times F}}$ ---- (2)
Here, M is the molecular mass of a substance and ${n_f}$ is the n-factor or number of electrons flowing and F is a Faraday constant.
Now substitute equation (2) in equation (1) to get:
$m = \dfrac{{M \times Q}}{{{n_f} \times F}}$
Now, we take M on the other side, so we get:
$\dfrac{m}{M} = \dfrac{Q}{{{n_f} \times F}}$
Now we know number of moles is ratio of given mass upon Molecular mass, so we can rewrite equation as:
$n = \dfrac{Q}{{{n_f} \times F}}$ ----(3)
Now we know from question:
Q=0.80 F
And we can write equation for Platinum ion converting into Platinum metal as below:
$P{t^{ + 4}} + 4{e^ - } \to Pt$
Therefore, we can write, ${n_f}=4$ (as 4 electrons are involved).
Now substitute the values in equation 3, we get:
$n = \dfrac{{0.8F}}{{4 \times F}}$
We cancel F from above and below and simplify to get:
$n = 0.2moles$
Thus, 0.2 moles of Platinum is deposited on Cathode.

Hence, the correct option is (C) 0.20 mol

Note: We know that the solution is of 1 molar, so 1 mole $P{t^{4 + }}$ will get deposited into 1 mole Platinum solid and 4 moles of electrons are used. No need to substitute the value of Faraday constant, as it gets cancelled. And, derive the equation for moles from the equation of mass deposited. Be careful regarding the value of Z, and substitute properly.