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Calculate the value of K given the following information anodel:(oxidation) Fe(s) Fe2+ (aq) + 2e E...
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С
A voltaic cell contains the following species involved in oxidation and reduction reactions: Au (s), Fe (s), Aut (aq), and Fe2+ (aq). What is the Eº of that cell? Half reaction Ed/V Aut(aq) +e Au(s) 1.692 Fe (aq) + 2e-> Fe(s) -0.447 2.937 V C 1.9155 V C 1.245 V 2.139 V 3.831 V
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
Consider the following half-reactions: Half-reaction E° (V) Ag+(aq) +e → Ag(s) 0.799V Cu2+(aq) + 2e → Cu(s) 0.337V Mg2+(aq) + 2e —— Mg(s) -2.370V The strongest oxidizing agent is: enter symbol The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will Mg(s) oxidize Ag(s) to Ag+(aq)? V Which species can be oxidized by Cu2+(aq)? If none enter none. Consider the following half-reactions: Half-reaction E° (V) 2Br (aq) 1.080V Br2(1) + 2e — Ni2+(aq) +...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
Table 20.2 Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr(s) -0.74 Fe2+ (aq) + 2e- Fe (5) -0.440 Fe3+ (aq) + e - Fe2+ (s) +0.771 Sn4+ (aq) + 2e- Sn2+ (aq) +0.154 The standard cell potential (Eºcell) for the voltaic cell based on the reaction below is V. 35n4+ (aq) + 2Cr (s) → 2Cr3+ (aq) + 3Sn2+ (aq) A) +1.94 B) +0.89 C) +2.53 D) -0.59 E) -1.02
NO−3(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) E∘=0.96V ClO2(g)+e−→ClO−2(aq) E∘=0.95V Cu2+(aq)+2e−→Cu(s) E∘=0.34V 2H+(aq)+2e−→H2(g) E∘=0.00V Pb2+(aq)+2e−→Pb(s) E∘=−0.13V Fe2+(aq)+2e−→Fe(s) E∘=−0.45V You may want to reference (Pages 898 - 902) Section 19.4 while completing this problem. Part A Use data from the table above to calculate E∘cell for the reaction. Fe(s)+2H+(aq)→Fe2+(aq)+H2(g) Express your answer using two decimal places.
Calculate the Ecell value at 298 K for the cell based on the reaction: Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) where [Ag+] = 0.00350 M and [Cu2+] = 7.00x10-4 M. The standard reduction potentials are shown below: Ag+(aq) +e → Ag(s) E° = 0.7996 V Cu2+ (aq) + 2e -→ Cu(s) E° = 0.3419 V 2nd attempt Ecell = V
4. Write a balanced equation from each cell notation, then calculate thee of the cell, is it spontaneous or nonspontaneous? A. Cr(s) Crot || Cui Cu(s) B. Al(s) Al* | Ce Ce Pt C. Cu (s) Cu? || Al Al (s) Useful information: F= 96485 J/V mol electrons, AG ----F-Ecell AG --RTINK: Ece 0.0592/n*logK; EcellEcett -0.0592/n*logQ Ce (aq) + Ag+ (aq) + e Fet (aq) + Cu" (aq) + Cu (aq) + 2e 2H(aq) + 2 Pb (aq) + 2e...