Complete Theory

5

Worked Examples

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

Exercises with Solutions

3
Exercise 1Standard cell potentialMedium
Problem to solve
Calculate Ecell∘E^\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 Zn∣Zn2+(x M)∣∣Cu2+(0.50 M)∣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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