Find k for this redox reaction when e = -2.12 v: zn2+ + 2cl- → zn + cl2.

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 = 2 × 10⁻⁷²

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole