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