The standard voltage of the cell [latex]Zn(s)|ZnSO_{4}(aq)||CuSO_{4}(aq)|Cu(s)[/latex] at 25.0°C is [latex]E^{\circ}_{cell} = 1.10\ V[/latex]. Given that [latex]E^{\circ}_{red} = -0.76\ V[/latex] for the electrode half reaction described by [latex]Zn^{2+}(aq, 1\ M) + 2\ e^− → Zn(s)[/latex] calculate [latex]E^{\circ}_{red}[/latex] at 25.0°C for the electrode half reaction given by [latex]Cu^{2+}(aq, 1\ M) + 2\ e^− → Cu(s)[/latex]

The oxidation takes place at the left-hand electrode, so the half reaction equations are [latex]Zn(s) → Zn^{2+}(aq) + 2\ e^−[/latex] (oxidation) [latex]Cu^{2+}(aq) + 2\ e^− → Cu(s)[/latex] (r eduction ) Because [latex]E^{\circ}_{ox}[Zn|Zn^{2+}] = –E^{\circ}_{red}[Zn|Zn^{2+}] = +0.76\ V[/latex] we have [latex]E^{\circ}_{cell} = E^{\circ}_{red}[Cu^{2+}|Cu] + E^{\circ}_{ox}[Zn|Zn^{2+}] = E^{\circ}_{red}[Cu^{2+}|Cu] + 0.76\ V = 1.10\ V[/latex] and so [latex]E^{\circ}_{red}[Cu^{2+}|Cu] = 1.10\ V – 0.76\ V = +0.34\ V[/latex] Thus, the standard reduction voltage of the [latex]Cu^{2+}(aq)|Cu(s)[/latex] electrode is [latex]E^{\circ}_{red} = +0.34\ V[/latex] at 25.0°C, in agreement with the value listed in Table 25.3. Therefore, the standard reduction voltage of an electrode can be obtained from the standard cell voltage of a cell for which the standard reduction voltage of the other electrode is known.

Question 25.7: The standard voltage of the cell Zn(s)|ZnSO4(aq)||CuSO4(aq)|......

General Chemistry [1096620]

The standard voltage of the cell

Zn(s)|ZnSO_{4}(aq)||CuSO_{4}(aq)|Cu(s)

at 25.0°C is $E^{\circ}_{cell} = 1.10\ V$ . Given that $E^{\circ}_{red} = -0.76\ V$ for the electrode half reaction described by

Zn^{2+}(aq, 1\ M) + 2\ e^− → Zn(s)

calculate $E^{\circ}_{red}$ at 25.0°C for the electrode half reaction given by

Cu^{2+}(aq, 1\ M) + 2\ e^− → Cu(s)

Step-by-Step

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The oxidation takes place at the left-hand electrode, so the half reaction equations are

$Zn(s) → Zn^{2+}(aq) + 2\ e^−$ (oxidation)

$Cu^{2+}(aq) + 2\ e^− → Cu(s)$ (reduction)

Because

E^{\circ}_{ox}[Zn|Zn^{2+}] = –E^{\circ}_{red}[Zn|Zn^{2+}] = +0.76\ V

we have

E^{\circ}_{cell} = E^{\circ}_{red}[Cu^{2+}|Cu] + E^{\circ}_{ox}[Zn|Zn^{2+}] = E^{\circ}_{red}[Cu^{2+}|Cu] + 0.76\ V = 1.10\ V

and so

E^{\circ}_{red}[Cu^{2+}|Cu] = 1.10\ V – 0.76\ V = +0.34\ V

Thus, the standard reduction voltage of the $Cu^{2+}(aq)|Cu(s)$ electrode is $E^{\circ}_{red} = +0.34\ V$ at 25.0°C, in agreement with the value listed in Table 25.3. Therefore, the standard reduction voltage of an electrode can be obtained from the standard cell voltage of a cell for which the standard reduction voltage of the other electrode is known.

TABLE 25.3 Standard reduction voltages at 25.0°C for aqueous solutions (see also Aِِppendix G)*
	Electrode half reaction	$E^{\circ}_{red}/V$
$\uparrow$ increasing strength of oxidizing agents	Acidic solutions		increasing strength of reducing agents $\downarrow$
	$F_{2}(g) + 2\ e^− → 2\ F^−(aq)$	+2.866
	$O_{3}(g) + 2\ H^+(aq) + 2\ e^− → O_{2}(g) + H_2O(l )$	+2.076
	$Co^{3+}(aq) + e^− → Co^{2+}(aq)$	+1.92
	$Cl_{2}(g) + 2\ e^− → 2\ Cl^−(aq)$	+1.358
	$O_{2}(g) + 4\ H^+(aq) + 4\ e^− → 2\ H_2O(l )$	+1.229
	$Pt^{2+}(aq) + 2\ e^– → Pt(s)$	+1.18
	$NO_{3}^{–}(aq) + 4\ H^+(aq) + 3\ e^– → NO(g) + 2\ H_2O(l )$	+0.957
	$Ag^+(aq) + e^- \to Ag(s)$	+0.7996
	$Cu^+(aq) + e^- \to Cu(s)$	+0.521
	$Cu^{+2}(aq) +2\ e^− → Cu(s)$	+0.342
	$Hg_2Cl_2(s) + 2\ e^− → 2\ Hg(l ) + 2\ Cl^−(aq)$	+0.268
	$AgCl(s) + e^- \to Ag(s) + Cl^-(aq)$	+0.2223
	$Cu^{2+}(aq) + e^− → Cu^+(aq)$	+0.153
	$2\ H^+(aq) + 2\ e^− → H_2(g)$	+0.0
	$Pb^{2+}(aq) + 2\ e^− → Pb(s)$	-0.126
	$V^{3+}(aq) + e^− → V^{2+}(aq)$	-0.255
	$Fe^{2+}(aq) + 2\ e^– → Fe(s)$	–0.447
	$Zn^{2+}(aq) + 2\ e^− → Zn(s)$	-0.762
	$Mn^{2+}(aq) + 2\ e^– → Mn(s)$	–1.185
	$Al^{3+}(aq) + 3\ e^− → Al(s)$	-1.662
	$H_2(g) + 2\ e^− → 2\ H^−(aq)$	-2.23
	$Mg^{2+}(aq) + 2\ e^− → Mg(s)$	-2.372
	$Na^+(aq) + e^- \to Na(s)$	-2.71
	$Ca^{2+}(aq) + 2\ e^– → Ca(s)$	–2.868
	$K^+(aq) + e^- \to K(s)$	–2.931
	$Li^+(aq) + e^- \to Li(s)$	-3.0401
	Basic solutions
	$O_2(g) + 2\ H_2O(l ) + 4\ e^− → 4\ OH^−(aq)$	+0.401
	$Cu(OH)_2(s) + 2\ e^− → Cu(s) + 2\ OH^−(aq)$	-0.222
	$Fe(OH)_3(s) + e^- \to Fe(OH)_2(s) + OH^-(aq)$	–0.56
	$2\ H_2O(l ) + 2\ e^− → H_2(g) + 2\ OH^−(aq)$	-0.8277
	$2\ SO_{3}^{2−}(aq) + 2\ H_2O(l) + 2\ e^− → S_{2}O_{4}^{2−}(aq) + 4\ OH^−(aq)$	-1.12
*Data from CRC Handbook of Chemistry and Physics, 87th ed., ed. David R. Lide, CRC Press, 2006–2007