Which of the following reactions occur at the negative electrode if a 1 M NaCl (aq) solution is used as a plating solution? Select a balanced reduction half-reaction below for your answer. Hint - Na metal cannot be produced in aqueous solutions. Na+ + e- → Na (s) Cl2 + 2 e- → 2 Cl- (aq) H2O + e- → 1/2 H2 (g) + OH- (aq) H2O → H+ (aq) + OH- (aq) Na + e- → Na- (aq)

(D) Na + e⁻ → Na⁻ (aq)

Explanation:

Negative electrode is cathode, and the type of reaction that occur here is reduction reaction.

In the electrolysis of aqueous solution of NaCl, the compound will dissociate into;

NaCl > Na⁺ + Cl⁻ and H₂O > H⁺ + OH⁻

At the negative electrode (cathode), we will have the following possible ions;

Na⁺ and H⁺

Hydrogen ion gains electron and hydrogen gas is produced;

H⁺ + e⁻ > H₂ (g)

Note: Sodium (Na) metal cannot be produced in aqueous solutions

Thus, Sodium metal will gain electron and become electronegative, which goes back into the solution. This is why it cannot be produced.

Na + e⁻ > Na⁻ (aq)

Therefore, the correct option is (D) Na + e⁻ → Na⁻ (aq)

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;