Question

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Nernst Equation Applied to Half-Reactions The Nernst equation can be applied to half-reactions. Calculate the reduction potential (at 25°C) of the half-cell Cu/Cu2+ (1.8x10- M). (The half-reaction is Cu2+ + 2e Cu.) 1 pts Submit Answer Tries 0/5 Calculate the reduction potential (at 25°C) of the half-cell Mn04 (2.00x10-1 M) Mn2+ (2.50X10-2 M) at pH = 4.00. (The half-reaction is MnO4 + 8H+ + 5e Mn2+ + 4H20.) 1pts Submit Answer Tries 0/5

Answer 1

aa NE 1) Given concentration of mat 2 1.8x10²M. Reaction - cut & re - aus Fred +0:34. The standard reduction potential of wilt to cu is +0.34 v. Using the vernst Egnation :- Fred & Ested - 2:30 3 RT log fant at 250, the verust equation es given as o- Ered & Ered – 0.059 - As cu is in 27 I solid state so, [(y) = 1) 2) Ered - 0.34- 11.88103 (no 2 as two • 0.34 - 0-0 295 log (055 x102) 2 0.34 - 0-0295 [log 0.55 + log 103] 2 0.34 – 0.0295 [-0.259 & 3] 2 0.34 - 0-0295 (2-74) ) a 0.34 - 0.080 = 0.4 V é are involved Thus, the reduction potential at 5c is 0.260

2) Given concentrations! - MnO4 = 2.00 x 10 m Mirt = 2.50 x 102M. & pur 4.00 Reaction is on Mun cag) + 8 H + (aq) + 5e The reduction potential for slus Mit of the le) Brd-H151 reaction can be given as - Cog [men 24] Ered & Ered - 0 [muo [ + 18 given, pra 4.00 tog [H + J = 4.00 & Leg (H f = – 4.00 & Det] - antilog (-4.00) (M+] = 0-0001 & H = 107 a. Ered & tored – 0.059 log (2.50 x10 2) 5 2.00x107) (104)8 ( no 5, as 5 e are involved & Ered = t 1.51 for the reation) Ered = 1.51 - 0-059 rog (1.25x1031) 5

Ered 1.51 -0.059 (log 1.25 + log 103 ) 21.51 -0.059 (0.096+ 31) 1.51 -0.059 ( 31-096) 2 1.51 - 0.366 - 1.14 V. Thus, the reduction potential at 250 for half cell es 1.14v. the