Consider the question posed in Exercise 6N.19 except that a saturated calomel electrode (the solution is saturated with KCl instead of having [Cl^–] = 1.00 mol · L^–1) is used in place of the standard calomel electrode. How will this replacement change your answers to Exercise 6N.19? The solubility of KCl is 35 g · (100 mL H₂O)^–1.

Exercise 6N.19

Suppose the reference electrode for Table 6M.1 were the standard calomel electrode, Hg₂Cl₂/Hg, Cl₂([Cl₂] = 1.00 mol · L^–1), with its E° set equal to 0. Under this system, what would be the potential for

(a) The standard hydrogen electrode;

(b) The standard Cu²⁺/Cu redox couple?

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TABLE 6M.1 Standard Potentials at 25 °C* Species Oxidized form is strongly oxidizing F₂/F Aut/Au Ce*+/Ce³+ MnO,H+/Mn²+ H₂O Cl₂/cl- Cr₂O,²,H+/Cr³+,H₂O 0₂.H™/H₂O Br₂/Br NO,,H+/NO,H₂O Ag+/Ag Fe³+/Fe²+ 1₂2/1 O2, H₂O/OH Cu²+/Cu AgCl/Ag,Cl- H+/H₂ Fe³+/Fe O₂,H₂O/HO₂ ,OH™ Pb²+/Pb Sn²+ /Sn Fe²+/Fe Zn²+/Zn H₂O/H₂,OH™ Al³+ /Al Mg²+/Mg Na+/Na K+/K Lit/Li Reduced form is strongly reducing Reduction half-reaction F₂(g) +2 e Au (aq) + e 2 F¯ (aq) Au(s) Ce+ (aq) + e Ce³+ (aq) MnO, (aq) + 8 H+(aq) + 5 eMn²+(aq) + 4 H₂O (1) Cl₂(g) + 2 e 2 Cl¯(aq) Cr₂O7² (aq) + 14 H(aq) + 6 e2 Cr³+(aq) + 7 H₂O(1) O₂(g) + 4 H (aq) + 4e 2 H₂O(1) Br₂(1) + 2 e2 Br (aq) NO, (aq) + 4 H+(aq) + 3 e Ag+ (aq) + e NO(g) + 2 H₂O(1) Ag(s) Fe³+(aq) +e→→ Fe²+ (aq) Iz(s) + 2 e 2I (aq) O₂(g) + 2 H₂O(1) + 4e4 OH (aq) →→→ Cu(s) 2+ Cu²+(aq) + 2 e AgCl(s) + e 2 H (aq) + 2 e Fe³+(aq) + 3 e Ag(s) + Cl (aq) H₂(g) Fe(s) O₂(g) + H₂O(1) + 2 eHO₂ (aq) + OH¯(aq) Pb²+ (aq) + 2 e→→→→ Pb(s) Sn²+ (aq) + 2 e Sn(s) Fe²+ (aq) + 2 e →→→→Fe(s) Zn²+ (aq) + 2 eZn(s) 2 H₂O(1) + 2 eH₂(g) + 2 OH (aq) Al³+ (aq) + 3 e- →→→→Al(s) Mg²+ (aq) + 2 e→→→→→→ Mg(s) Na (aq) + e→→→→→→→ Na(s) K+ (aq) + eK(s) Lit(aq) +eLi(s) +2.87 +1.69 +1.61 +1.51 Eº/V +1.36 +1.33 +1.23; +0.82 at pH = 7 +1.09 +0.96 +0.80 +0.77 +0.54 +0.40; +0.82 at pH = 7 +0.34 +0.22 0, by definition -0.04 -0.08 -0.13 -0.14 -0.44 -0.76 -0.83; -0.42 at pH = 7 -1.66 -2.36 -2.71 -2.93 -3.05