3) The voltaic cell described by the cell notation has an Eo of 0.68 V. Calculat
ID: 1003076 • Letter: 3
Question
3) The voltaic cell described by the cell notation has an Eo of 0.68 V. Calculate the Wmax (kJ) the cell has done if 6.4700 mol of MnO4-(aq) (Molar Mass - 118.94 g/mol) reacts. Round your answer to 3 significant figures.
Pt(s) l VO2+(aq), VO2+(aq), H+(aq) ll H+(aq), MnO4-(aq) l MnO2(s) l Pt(s)
St. Red. Pot. (V)
VO2+/VO2+: +1.00
MnO4-/MnO2 :+1.68
Faraday's Constant
F = 96485 C
4) The galvanic cell described by the cell notation has an Eo of 1.23 V. Calculate the value (kJ) for the G of the cell. Round your answer to 3 significant figures.
Pt(s), H2(g) l H2O(l), OH-(aq) ll O2(g) l OH-(aq) l Pt(s)
St. Red. Pot. (V)
H2O/H2: -0.83
O2/OH- : +0.40
Faraday's Constant
F = 96485 C
5) The electrochemical cell described by the balanced chemical equation has a standard cell potential of -0.37 V. Calculate the value (kJ) for the standard free energy change of the cell. Round your answer to 3 significant figures.
Sn4+(aq) + 2Cu(s) Sn2+(aq) + 2Cu+(aq)
St. Red. Pot. (V)
Cu+/Cu: +0.52
Sn4+/Sn2+: +0.15
Faraday's Constant
F = 96485 C
6)The electrochemical cell described by the cell notation has an Eocell of 4.05 V. Calculate the maximum electrical work (kJ) the cell has done if 375.44 g of F2(g) (Molar Mass - 38.00 g/mol) reacts. Round your answer to 3 significant figures.
Mn(s) l Mn2+(aq) ll F2(g) l F-(aq) l Pt(s)
St. Red. Pot. (V)
Mn2+/Mn: -1.18
F2/F- : +2.87
Faraday's Constant
F = 96485 C
Explanation / Answer
3) The voltaic cell described by the cell notation has an Eo of 0.68 V. Calculate the Wmax (kJ) the cell has done if 6.4700 mol of MnO4-(aq) (Molar Mass - 118.94 g/mol) reacts. Round your answer to 3 significant figures.
Pt(s) l VO2+(aq), VO2+(aq), H+(aq) ll H+(aq), MnO4-(aq) l MnO2(s) l Pt(s) St. Red. Pot. (V) VO2+/VO2+ +1.00 MnO4-/MnO2 +1.68 Faraday's Constant F = 96485 C
w(max) = -nFEo
n = number of moles of e- which is transferred in the reaction; 2
F = 1 Faraday = 96,485 Coulombs (C)
Eo =0.68 V
= 0.68 J/C
w(max) = -nFEo = -2 x 96,485 C x (0.68 J/C)
= 131219.6 J
= 131.22 KJ
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