CHEMISTRY HOMEWORK DUE THURSDAY EVENING. KINDLY SHOW ALL WORKINGS AND CALCULATIO

Question

CHEMISTRY HOMEWORK DUE THURSDAY EVENING. KINDLY SHOW ALL WORKINGS AND CALCULATIONS

(3). What is the value of the equilibrium constant for the cell reaction below at 25°C?

2Cr (s) + 3Pb2+ (aq) 3Pb (s) + 2Cr3+ (aq)        E°cell = 0.61 V

(4). Calculate Ecell at 25 oC for the following reaction under the conditions given. Note that you must first balance the equation in the problem. E°cell = 0.52 V

H2S (g) + NO3(aq) S (s) + NO (aq)       (in acidic solution)

[NO3-] = 1.6 M

p NO = 0.33 atm

p H2S = 0.9 atm

[H+] = 1.2 M

Explanation / Answer

E0cell = ( RT / nF ) lnK

ln K = nFE0cell / RT

ln K = 6 * 96500 * 0.61 / 8.314 * 298

ln K = 142.555

K = 8.14 * 1061

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