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poizon [28]
3 years ago
15

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]3 years ago
8 0

<em>K</em> = 2.4 × 10^(-72)

<em>Step 1</em>. Determine the <em>value of n </em>

Zn^(2+) + 2e^(-) → Zn

2Cl^(-) → Cl_2 + 2e^(-)

Zn^(2+) + 2Cl^(-) → Zn + Cl_2

∴ <em>n</em> = 2

<em>Step 2</em>. Calculate <em>K</em>

log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62

<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)

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