0 votes
3 views
ago in ElectroChemistry by (450 points)
Given the following half-reactions:Cu2+ (aq) + 2 e- -> Cu (s)    E° = +0.34 VZn2+ (aq) + 2 e- -> Zn (s)     E° = -0.76 VCalculate the equilibrium constant (K) for the reaction: Cu2+ (aq) + Zn (s) -> Zn2+ (aq) + Cu (s)

1 Answer

0 votes
ago by (610 points)
To calculate the equilibrium constant  K  for the given reaction, we first need to determine the standard cell potential  Ecell  for the reaction. The overall reaction can be obtained by combining the two half-reactions:1. Cu2+  aq  + 2 e- -> Cu  s     E = +0.34 V2. Zn  s  -> Zn2+  aq  + 2 e-    E = +0.76 V  reverse the second half-reaction Add the two half-reactions:Cu2+  aq  + Zn  s  -> Zn2+  aq  + Cu  s Now, add the standard potentials for the two half-reactions to get the standard cell potential  Ecell :Ecell = E Cu2+/Cu  + E Zn/Zn2+ Ecell = +0.34 V +  +0.76 V Ecell = +1.10 VNow, we can use the Nernst equation to calculate the equilibrium constant  K . The Nernst equation is:Ecell =  RT/nF  * ln K Where:Ecell = standard cell potential  +1.10 V R = gas constant  8.314 J/molK T = temperature  assuming 298 K n = number of electrons transferred  2 in this case F = Faraday's constant  96485 C/mol Rearrange the equation to solve for K:ln K  =  nF * Ecell  /  RT Now, plug in the values:ln K  =  2 * 96485 C/mol * 1.10 V  /  8.314 J/molK * 298 K ln K   50.71Now, take the exponential of both sides to solve for K:K  e^50.71 K  6.2  10^21The equilibrium constant  K  for the reaction Cu2+  aq  + Zn  s  -> Zn2+  aq  + Cu  s  is approximately 6.2  10^21.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...