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

Chemistry

1 answer:

anygoal [31]4 years ago

8 0

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)

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