First slide

Faraday's Laws

Question

In the electrochemical conversion (Kolbe’s eletrolysis) of R–COONa to R–R, 1A current was passed for 965 seconds. Calculate the amount of R–R formed in this process (Faraday constant = 96,500 C mol^–1)

Moderate

A

10m mol
B

5m mol
C

100m mol
D

50m mol

Solution

$2 RCOONa + 2 H_{2} O \overset{2 F}{⟶} \underset{Anode}{R -} R + 2 {CO}_{2} + \underset{Cathode}{2 NaOH + H_{2}}$
1 mole of R-R = 2 F = 2 x 96500 coulomb
No.of Coulombs consumed = i $\times$ t =1 x 965 =965 C
$No.of moles of R - R formed = \frac{965}{2 \times 96500} = \frac{1}{200} mol ; Number of milli moles = 5;$

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