Question

Silver nitrate solutions are often used to plate silver onto other metals. what is the maximum amount of silver (in grams) that can be plated out of 4.8 l of an agno3 solution containing 3.4% ag by mass? assume that the density of the solution is 1.01 g>ml

Answer 1

The maximum amount of silver (in grams) that can be plated out = 164,832 grams

Further explanation

Electroplating is an application of electrolysis. In the plating process, the metal to be plated as a cathode and metal plating function as an anode. The two electrodes were inserted into the electrolyte salt solution from the positive metal plating ion. In this process, the metal in the anode undergoes oxidation and dissolves into its metal ions. Then this ion will experience a reduction reaction and is deposited on the surface of the cathode.

Example of using silver nitrate (AgNO₃) solution in plating to iron. The silver electrode at the anode will be oxidized to Ag⁺ ions

At the cathode, a reduction in Ag⁺ ions from the electrolyte solution will be deposited on Fe (iron). As for the cathode, there will be a reduction reaction of silver ions to silver metal, the silver metal is deposited on the cathode (Fe). The process occurs until all silver ions are reduced so that Fe is coated by Ag

Reactions that occur:

Anode: Ag (s) → Ag⁺ (aq) + e⁻

Cathode: Ag⁺ (aq) + e⁻ → Ag (s)

-------------------------------------------------- ----

             Ag (anode) → Ag (cathode)

4.8 l of an AgNO₃ solution containing 3.4% Ag by mass

• We determine the mass of the solution

Density formula:

$\large{\boxed{\bold{\rho~=~\frac{m}{V} }}}$

mass of solution = density x volume

solution mass = 1.01 g/ ml (= 1 kg / l) x 4.8 L

mass of solution = 4,848 kg

• We determine the mass of Ag

Percentage of mass Ag in AgNO₃ = 3.4%

so the mass Ag:

mass Ag = 3.4% x 4,848 kg

mass Ag = 0.164832 kg

mass of Ag = 164,832 grams

Learn more

an electrolytic cell

brainly.com/question/1319774

During electrolysis

brainly.com/question/11868966

the substance is produced at the cathode during the electrolysis

brainly.com/question/6854891

Keywords: electrolysis, AgNO₃

Answer 2

Answer is: the maximum amount of silver is 1.528 mol.

V(solution) = 4.8 L · 1000 mL/L.
V(solution) = 4800 mL.
d(solution) = 1.01 g/ml.
m(solution) = d(solution) · V(AgNO₃).
m(solution) = 1.01 g/mL · 4800 mL.
m(solution) = 4848 g.
ω(Ag) = 3.4% ÷ 100%.
ω(Ag) =0.034.
m(Ag) = ω(Ag) · m(solution).
m(Ag) = 0.034 · 4848 g.
m(Ag) = 164.832 g; mass of silver.
n(Ag) = m(Ag) ÷ M(Ag).
n(Ag) = 164.832 g ÷ 107,8682 g/mol.
n(Ag) = 1.528 mol; amount of substance.