Answer: B, C, D
Silver deposited =5.40/108=0.0500 mol, so QA=0.0500×96500=4825 C and IA=4825/3860=1.25 A; (B) correct. In parallel IB=3.25−1.25=2.00 A, so QB=2.00×3860=7720 C=0.0800 mol of electrons; (C) correct. Copper =0.0400×63.5=2.54 g; (D) correct. In series the full 3.25×3860=12545 C=0.130 mol of electrons would pass through B giving 0.065×63.5=4.13 g Cu, not 8.26 g (which ignores n=2); (A) wrong.