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

the answer is c

tell me if wrong

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.