Estimate E cell for the half reaction. 2H 2 O + 2e - H 2 +
Question:
Estimate E°cell for the half reaction. 2H2O + 2e- → H2 + 2OH- given the following values of ΔGof :
H2O(l) = –237 kJ/ mol H2(g) = 0.0 OH-(aq) = –157 kJ/ mol e- = 0.0 Compare this value of E°cell with the value of E°cell given in Table.
Table
Transcribed Image Text:
Standard Reduction Potentials at 25°C (298 K) for Many Common Half-reactions g° (V) g° (V) Half-reaction Half-reaction O2 + 2H,0 + 4e - 40H- Cu2+ + 2e- - Cu Hg:Cl2 + 2e- AGCI + e- → Ag + Cl- So- + 4H* + 2e- → H,SO, + H20 Cu2+ + e- → Cu* 2H* + 2e Fel+ + 3e- Pb2+ + 2e- Sn2+ + 2e- Ni2* + 2e- → Ni PBSO, + 2e- - Pb + SO2- Cd2+ + 2e- Fe+ + 2e" Cr3+ + e Cr+ Cr+ + 3e- - Cr Zn2+ + 2e Zn 2H;0 + 2e- Mn2+ + 2e" - Mn Al3+ + 3e Al H2 + 2e- Mg+ + 2e → Mg La3+ + 3e- La F2 + 2e- 2F- Ag+ + e Ag* Co+ + e Co+ H2O2 + 2H* + 2e" 2H20 Ce+ + e - Ce+ PbO, + 4H* + SO2- + 2e- - PbSO, + 2H20 Mno,- + 4H* + 3e- → MnO, + 2H;0 10,- + 2H+ + 2e- I0;- + H20 MnO,- + 8H* + Se + Mn2+ + 4H;0 Au3+ + 3e- - Au PbOz + 4H* + 2e- - Pb?+ + 2H20 Cl, + 2e- Cr,02- + 14H + 6e- - O2 + 4H+ + 4e- + 2H,0 MnO, + 4H+ + 2e IO;- + 6H* + Se → H2 + 3H20 Brz + 2e- + 2Br- VO,* + 2H* + e VO2+ + H,0 AuCl, + 3e" - Au + 4CI- NO;- + 4H* + 3e- - NO + 2H20 CIO, + e- → CIO,- 2Hg+ + 2e-- Ag* + e + Ag Hgz2+ + 2e- - 2Hg Fe* + e - Fe+ O, + 2H* + 2e" - H2O2 MnO,- + e- → Mno,- I+ 2e → 21- Cu* + e- → Cu 2.87 0.40 0.34 1.99 - 2Hg + 2C1- 1.82 1.78 1.70 1.69 1.68 1.60 0.27 0.22 0.20 0.16 0.00 -0.036 Н Fe -0.13 -0.14 1.51 1.50 Pb Sn -0.23 1.46 1.36 1.33 1.23 + 201- -0.35 -0.40 2Cr* + 7H,0 - Cd Fe -0.44 Mn2+ + 2H;0 -0.50 1.21 1.20 1.09 1.00 -0.73 -0,76 H2 + 20H- -0.83 0.99 0.96 0.954 -1.18 -1.66 -2.23 + 2H- Hg;+ 0.91 -2.37 -2.37 0.80 0.80 Na Na -2.71 -2.76 -2.90 Ca2+ + 2e- - Ca Ba?+ + 2e" 0.77 0.68 - Ba 0.56 0,54 0.52 K* + e- → K -2.92 Lit +e → Li -3.05
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