📋 Formulario

Complete Theory

Worked Examples

Example 1Cell EMF with Nernst equation
Example 2Mass deposited by electrolysis

Exercises with Solutions

Exercise 1Standard cell potentialMedium
📋 Problem to solve
Calculate EcellE^\circ_{cell} for Mg(s)+2Ag+(aq)Mg2+(aq)+2Ag(s)\mathrm{Mg(s) + 2Ag^+(aq) \to Mg^{2+}(aq) + 2Ag(s)}. E(Mg2+/Mg)=2.37E^\circ(\mathrm{Mg^{2+}/Mg}) = -2.37 V, E(Ag+/Ag)=+0.80E^\circ(\mathrm{Ag^+/Ag}) = +0.80 V.
📌 Given data
E°(Mg²⁺/Mg) = −2.37 VE°(Ag⁺/Ag) = +0.80 V
Exercise 2Nernst equationMedium
📋 Problem to solve
For the cell ZnZn2+(xM)Cu2+(0.50M)Cu\mathrm{Zn|Zn^{2+}(x\,M)||Cu^{2+}(0.50\,M)|Cu}, Ecell=1.12E_{cell} = 1.12 V. Find [Zn2+][\mathrm{Zn^{2+}}]. Ecell=1.10E^\circ_{cell} = 1.10 V.
📌 Given data
Ecell=1.12E_{cell} = 1.12 VEcell=1.10E^\circ_{cell} = 1.10 V[Cu2+]=0.50[\mathrm{Cu^{2+}}] = 0.50 Mn=2n = 2T = 25°C
Exercise 3Electrolysis timeHard
📋 Problem to solve
How long must a current of 5.00 A be passed to deposit 10.0 g of silver from AgNO₃? M(Ag)=107.87M(\mathrm{Ag}) = 107.87 g/mol.
📌 Given data
I = 5.00 Am = 10.0 gM = 107.87 g/molAg⁺ + e⁻ → Ag (n=1)F = 96485 C/mol
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