What is the value of E degree and Delta G degree for the following electrochemic

Question

What is the value of E degree and Delta G degree for the following electrochemical reaction? 10CI^- (aq) + 2MnO_4^- (aq) + 16H^+ (aq) rightarrow 2Mn^2+ (aq) + 8H_2 O_(I) + 5CI_2 (g)? MnO_4^- (aq) + 8H^+ (aq) + 5e^- rightarrow Mn^2+ (aq) + 4H_2 O_(I) E degree = +1.51 V CI_2 (g) + 2e^- (aq) rightarrow 2CI^- (aq) E degree = +1.36V E degree Delta G degree A. -2.87 V 2.77 times 103 kJ/mol B. -0.15 V 145 kJ/mol C. 0.15 V -145 kJ/mol D. 2.87 V -2.77 times 103 kJ/mol E. All the above F. None of the above

Explanation / Answer

Given reactions are

MnO4- + 8H+ +5e- -->Mn+2+4H2O,   Eo=1.51V (1)

Cl2+2e- ------->2Cl-, Eo= 1.36V   or 2Cl- ------>Cl2+2e-, Eo=-1.36V (2)

Multiplying Eq.2 with 4 and Eq.1 with 2 gives the desired reaction

2MnO4- +16H++10Cl- ----->2Mn+2+8H2O+5Cl2, Eo=1.51-1.36=0.15V

deltaGo=-n ( no of electrons transferred) *FE= -10*96500*0.15 =-144750 J/mole= -145KJ/mole ( C is correct)

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