What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr₂(SO₄)₃ solution using a current of 0.5 A
Cr³⁺ + 3e⁻ → Cr . 1 mol Cr (52 g) requires 3F. Moles = (0.585/52) = 0.01125 mol , Charge = 0.01125 × 3 × 96500 = 3256.875 C . t = (Q/I) = (3256.875/0.5) = 6513.75 s .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential