The standard cell potential (E degree cell) of the reaction below is +0.126 V. W

Question

The standard cell potential (E degree cell) of the reaction below is +0.126 V. What is the value of delta G degree for this reaction (kJ/mol) at 25 degree C? Pb (s) + 2H^+(aq) rightarrow Pb^2+ (aq) + H_2 (g) -24.3 +24.3 -12.6 +12.6 The standard cell potential (E degree cell) of the reaction below is -0.55 V. What is the equilibrium constant K for this reaction at 25 degree C? I_2 (s) + 2Br^- (aq) rightarrow 2I^- (aq) + Br_2 (l) 4.9 times 10^-21 K 2.5 times 10^-19 5.0 times 10^-10 4.0 times 10^18

Explanation / Answer

Solution :-

Q37 ) Eo cell = +0.126 V

Delta Go = ?

T= 25 C +273 = 298 K

Pb(s) + 2H^+(aq)   ----- > Pb^2+(aq) + H2(g)

Formula

Delta Go = - n*F*Eo cell

                = -2 * 96500 C * 0.126

              = -24318 J/mol

-24318 J per mol * 1 kJ/ 1000 J = -24.3 kJ/mol

So the answer is option A that is -24.3 kJ/mol

Q38 ) I2 + 2Br^-    ---- > Br2 + 2I^-

Eo cell = -0.55 V

T= 25 C +273 = 298 K

Equilibrium constant K = ?

Eo cell = [RT/nF]* ln K

Eo cell / [RT/nF] = ln K

-0.55 V / [8.314 J per mol K * 298 K / 2 * 96500 C] = ln K

-42.84 = ln K

Anti ln [-42.84] = K

2.5 *10^-19 = K

So the answer is option B

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