ElectrochemistryHard
Question
From an electrolyte, one mole of electron will deposit at cathode
Options
A.63.5 g of Cu
B.24 g of Mg
C.11.5 g of Na
D.9.0 g of Al
Solution
$n_{eq}Cu = \frac{63.5}{63.5} \times 2 = 2;n_{eq}Mg = \frac{24}{24} \times 2 = 2$
$n_{eq}Na = \frac{11.5}{23} \times 1 = 0.5;n_{eq}Al = \frac{9}{27} \times 3 = 1$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
The standard reduction potentials of Pt|Cr2O72−, Cr+3; Pt|MnO4−, Mn+2; Pt|Ce+4, Ce+3 in the presence of acid are 1.33 V,...During the preparation of H2S2O8 (per disulphuric acid) O2 gas also releases at anode as byproduct, When 9.72 L of H2 re...Estimate the cell potential of a Daniel cell having 1.0 M – Zn2+ and originally having 1.0 M – Cu2+ after sufficient amm...In the electrochemical cell :- Zn|ZnSO4(0.01M)| |CuSO4(1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentrat...The quantity of electricity required for the reduction of 1 mole of Fe2O3 to Fe is...