Given: logK=nE∘0.0592 What is the value K for this redox reaction? Zn2 + (aq) + 2 Cl - (aq) → Zn (s) + Cl2 (

g. Ecell = - 2.12 V

K = 2.4 * 10^ (-72)

Step 1. Determine the value of n

Zn^ (2+) + 2e^ (-) → Zn

2Cl^ (-) → Cl_2 + 2e^ (-)

Zn^ (2+) + 2Cl^ (-) → Zn + Cl_2

∴ n = 2

Step 2. Calculate K

logK = nE°/0.0592 V = [2 * (-2.12 V) ]/0.0592 V = - 71.62

K = 10^ (-71.62) = 2.4 * 10^ (-72)