An electric current of 0.965 ampere is passed for 2000 s through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposits d at the cathode. The amount of Cu deposited is

Question

An electric current of 0.965 ampere is passed for 2000 s through a solution containing [Cu(CH3CN)4]+ and metallic copper is deposits d at the cathode. The amount of Cu deposited is (A) 0.005 mol (B) 0.01 mol (C) 0.02 mol (D) 0.04 mol

Tardigrade Admin · Accepted Answer

[Cu(CH3CN)4]++e- xrightarrow[]Cu+4CH3CN For the above reaction 1 F charge is required to deposit 1 mole of Cu (i.e. 63.5 g) Charge supplied =0.965× 2000 =2× 965=0.02 text F ∴ Moles of Cu deposited = 0.02 mol

Q. An electric current of 0.965 ampere is passed for 2000 s through a solution containing $[C u (C H_{3} CN)_{4}]^{+}$ and metallic copper is deposits d at the cathode. The amount of Cu deposited is

0.005 mol

0.01 mol

0.02 mol

0.04 mol

$[C u (C H_{3} CN)_{4}]^{+} + e^{-} C u + 4 C H_{3} CN$ For the above reaction 1 F charge is required to deposit 1 mole of Cu (i.e. 63.5 g) Charge supplied $= 0.965 \times 2000$ $= 2 \times 965 = 0.02 F$ $∴$ Moles of Cu deposited = 0.02 mol

Q. An electric current of 0.965 ampere is passed for 2000 s through a solution containing [Cu(CH3​CN)4​]+ and metallic copper is deposits d at the cathode. The amount of Cu deposited is