Question

Find K for this redox reaction when E = -2.12 V: Zn2+ + 2Cl- → Zn + Cl2.

Answer 1

K = 2 × 10⁻⁷²

Explanation:

Step 1. Determine the value of n

Zn²⁺ + 2e⁻ → Zn

2Cl⁻ → Cl₂+ 2e⁻

Zn²⁺ + 2Cl⁻ → Zn + Cl₂; E = -2.12 V

n = 2 mol electrons

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Step 2. Calculate K

The formula relating E and K is

E = (RT)/(nF)lnK

E = -2.12 V

R = 8.314 J·K⁻¹mol⁻¹

T = 298.15 K

n = 2

F = 96 485 C/mol

-2.12 = (8.314 × 298.15)/(2 × 96 485)lnK Do the multiplications

-2.12 = 2479/192 970 lnK Do the division

-2.12 = 0.012 85 lnK Divide both sides by 0.012 85 and transpose

ln K = -165.0 Take the antilog of both sides

`K = e^(-165.0)`

K = 2 × 10⁻⁷²