What is the equilibrium constant (K) at 25 degree C for the following cell react

Question

What is the equilibrium constant (K) at 25 degree C for the following cell reaction? Zn(s) + cr^2+(aq) rightarrow cr(s) + zn^2+ (aq) E = 0.89 V A) 3.7 times 10^12 B) 1.8 times 10^-24 C) 1.2 times 10^ D) 4.5 times 10^ E) 3.7 times 10^ what is E of the following cell reaction at 25 degree E = 0.460 v. A) 0.426 V B) 0.505 V C) 0.559 V D) 0.538 V E) 0.463 V The rate constants for the first-order decomposition of a compound are 6.80 times 10^-4 s^-1 at 45 degree C and 4.22 times 10^-3 s^-1 at 122 degree C. What is the value of the activation energy for this reaction? (R = 8.31 J/(mol middot K)) A) 71.4 kJ/mol B) 40.6 kJ/mol C) 0.277 kJ/mol D) 24.2 kJ/mol E) 0.896 kJ/mol

Explanation / Answer

12)

here, number of electrons being transferred, n = 2

E = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.0592

So, E = (0.059/n)*log Kc

0.89 = (0.0592/2)*log Kc

log Kc = 30.068

Kc = 1.2*10^30

Answer: 1.2*10^30

13)

Balance chemical equation is:

Cu (s) + 2Ag+ (aq) —> Cu2+ (aq) + 2Ag (s)

Number of electron being transferred in balanced reaction is 2

So, n = 2

Use:

E = Eo - (2.303*RT/nF) log {[Cu2+]^1/[Ag+]^2}

2.303*R*T/n = 2.303*8.314*298.0/F= 0.0591

E = Eo - (0.0591/n) log {[Cu2+]^1/[Ag+]^2}

E = 0.46 - (0.0591/2) log (0.65^1/0.92^2)

E = 0.46-(-0.003)

E = 0.463 V

Answer: E

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