Question

The amount of charge required to reduce 1 mole \[Cr_{2}O_{7}^{--}\] ion in acidified medium is:
a)28950 C
b)43425 C
c)57900 C
d)86850 C

Answer 1

@Ujjwal!
Write down the half-equation from \(Cr_2O_7\:^{2-}\) to \(Cr^{3+}\) to find the number n of electrons exchanged.

Then, amount of charge will be n.F, where F is Faraday's constant, roughly F = 96500 C/mol or 96485 C/mol

These 96500 coulombs represent one mole of elementary charge.
F = \(N_A\,e\) = \(6.022\:10^{23} \times 1.602\:10^{-19}\)

http://en.wikipedia.org/wiki/Faraday%27s_constant

Answer 2

Since n = 6, then 6 x 96500 = 579,000 C

There is a 0 missing in the answer

Answer 3

The question mentions 'charge required to reduce'.. But in your answer, i can see it being oxidized..

Answer 4

\(Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O\)

Dichromate ion gains electrons so it is reduced.

Answer 5

I get it.. Thanks. But how do i find out that equation?

Answer 6

what is faradays constant,i didnt get it

Answer 7

@bhusal
Faraday's constant is the magnitude of electric charge per mole of electrons..

http://en.wikipedia.org/wiki/Faraday_constant

Answer 8

@bhusal
Follow the link given in my previous message.