On passing electric current of one ampere for 16 min and 5 s through one litre solution of CuCl2 , all copper solution was deposited at cathode. The strength of CuCl2 solution was (molar mass of Cu = 63.5, faraday constant = 96500 C/mol)

Question

On passing electric current of one ampere for 16 min and 5 s through one litre solution of CuCl2 , all copper solution was deposited at cathode. The strength of CuCl2 solution was (molar mass of Cu = 63.5, faraday constant = 96500 C/mol) (A) 0.2 N (B) 0.01 N (C) 0.001 N (D) 0.02 N

Tardigrade Admin · Accepted Answer

Key Idea According to Faraday's first law

W = Z I t

where,

W =

mass of substance deposited (gram)

I =

current (amperes)

t =

time

(

second

)

Z =

electrochemical equivalent
and

Z = \frac{E}{96500}

where,

E =

equivalent weight

E = \frac{atomic weight}{valency}

Here,

I = 1 A

,
Molar mass of copper

= 63.5

t = 16 min

and

5 s

= 16 \times 60 + 5 = 965 s

Using formula

m = \frac{E}{96500} \times I \times t

= \frac{63.5/2}{96500} \times 1 \times 965

= \frac{63.5}{200} = 0.3175 g

Volume of

C u C l_{2}

solution

= 1 L

∵

Normality

= \frac{weight of solute/eq. wt. of solute}{volume of solution (in litre)}

∴

Normality of

C u C l_{2}

solution

= \frac{0.3175}{31.75 \times 1}

= 0.01 N

Q. On passing electric current of one ampere for 16 min and 5s through one litre solution of CuCl2​ , all copper solution was deposited at cathode. The strength of CuCl2​ solution was (molar mass of Cu=63.5, faraday constant =96500C/mol)

0.2 N

0.01 N

0.001 N

0.02 N

Q. On passing electric current of one ampere for $16$ min and $5 s$ through one litre solution of $C u C l_{2}$ , all copper solution was deposited at cathode. The strength of $C u C l_{2}$ solution was (molar mass of $C u = 63.5$ , faraday constant $= 96500 C / m o l)$