21. i thought understood electrolysis but i guess not. can someone please explai

Question

21. i thought understood electrolysis but i guess not. can someone please explain why I2 is the answer?

as far as e calculations i am confused as to when i change the sign to negative and which one i do it to when they are both set in reduction already. do i use the most likely to oxidize as my anode? so 0.77-1.51?

21 In the electrolysis of a molten mixture of KIand MgF, identity the product that forms at the D) 22. Calculate AG for the reaction of iron(1) ions with one mole of permanganate ions. Half reaction 151 V 077 V A) -714 C) -357 kJ DO-428 23. Calculate Ea for the reaction of nickel ons with cadmium metal at 25 C. K 1.17 x A)0.075 V B) 0.10 V 0.30 V D) 0.15 V 24. Considering that the E for iron equals -0A4 V. Which of the following metals can be used as a sacrificial anode to protect against c underground iron pipes: Mg (E 2.37

Explanation / Answer

In the electrolysis of mixture of molten salts, cation with higher Eo value i.e. stronger oxidising agent gets reduced at cathode and anion with lower Eo value i.e. stronger reducing agent gets oxidised at anode.

Thus, in the given molten mixture of KI and MgF2, there are two cations, K+ (Eo = -2.925 ) and Mg2+ (Eo = -2.37). Mg2+ ion being stronger oxidsing agent (higher Eo), has greater tendency to get reduced at cathode and Mg is deposited at cathode.

On the other hand, out of two anions, I- (Eo = 0.535) and F- (Eo = 2.87), I- ion being strong reducing agent (lower Eo), has greater tendency to get oxidised at anode.Therefore, I2 is formed at anode.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.