Dear Student, ∆G=∆G°+ RTln Kor, ∆G°=-RTlnK+∆Galso,∆G=-nFEcellSo, ∆G°=-RTlnK-nFEcellKf=XSo, ∆G°=-RTlnX-nFEcellFor the reaction, Au + 2CN- + 14O2 + 12H2O → Au(CN)2- + OH-Here Au is converting to Au+So, Au3+ + 3e- → Au (i)∆G°=-nFE°=-3F×1 5=-4 5FAu3+ + 2e- → Au+ (ii)∆G°=-nFE°=-2F× 1 4=-2 8FSubstracting (ii) from (i), we get the desired equationi e Au+ + e-→Au (iii)So, ∆G°= -4 5F-(-2 8F) = -4 5F+2 8F=-1 7Fand, E°=-nF∆G°=-1×F-1 7F=0 5275VE°cell = E°cathode-E°anode =-0 5275V-0 41V =-0 9375 VSo, Ecell = E°cell=-0 9375Vor, ∆G =-nFEcell =-4×F×(-0 9375) =2 11 For, ∆G°=-RTlnX-2 11 FSo, correct answer is (B)

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Please help asap Find (in Q-t for strong ABZ in water - 0.25 Conc. C ( mole/L) 160 150 180 A" + 2CN- (aq.' + —O, (g) + — + OH- Use the following data to calculate AG' for the reaction E' = + 041 Au Are • + 2e— RT X. 1 29 F (C) -RT/n—• A hydrogen electrode X was placed an a buffer solution Of ac a b and another elee•ttode Y place'! tn

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∆ G = ∆ G ° + RTln K or, ∆ G ° = - RTlnK + ∆ G also, ∆ G = - {nFE}_{cell} So, ∆ G ° = - RTlnK - {nFE}_{cell} K_{f} = X So, ∆ G ° = - RTlnX - {nFE}_{cell} For the reaction, Au + 2 {CN}^{-} + \frac{1}{4} O_{2} + \frac{1}{2} H_{2} O \to Au (CN {)_{2}}^{-} + {OH}^{-} Here Au is converting to {Au}^{+} So, {Au}^{3 +} + 3 e^{-} \to Au . . . (i) ∆ G ° = - nFE ° = - 3 F \times 1.5 = - 4.5 F {Au}^{3 +} + 2 e^{-} \to {Au}^{+} . . . (ii) ∆ G ° = - nFE ° = - 2 F \times 1.4 = - 2.8 F Substracting (ii) from (i), we get the desired equation i . e . {Au}^{+} + e^{-} \to Au . . . (iii) So, ∆ G ° = - 4.5 F - (- 2.8 F) = - 4.5 F + 2.8 F = - 1.7 F and, E ° = \frac{- nF}{∆ G °} = \frac{- 1 \times F}{- 1.7 F} = 0.5275 V E °_{cell} = E °_{cathode} - E °_{anode} = - 0.5275 V - 0.41 V = - 0.9375 V So, E_{cell} = E °_{cell} = - 0.9375 V or, ∆ G = - {nFE}_{cell} = - 4 \times F \times (- 0.9375) = 2.11 F or, ∆ G ° = - RTlnX - 2.11 F So, correct answer is (B)

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