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)

explanation:

i think option no d is the correct answer

the answer is   c.) 1575 j.

explanation: