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jcherry99

Answer:

Explanation:

Trick question. The cathode is where the reduction reaction takes place. The reduction reaction is the gain of electrons.

Pb+2 + 2e^- ===> Pb        The eo for that is - 0.126.

The minus sign indicates that the Pb^2+ is not overjoyed at taking on those two electrons. If it had a say in the matter, it would rather be giving up electrons. In other words, it would rather be the oxidizing equation which would look like this

Pb ===> Pb+2 + 2e^- and the oxidizing potential would be eo = + 0.126

That's what moving right and moving left means. If the eo is - then the preferred reaction is the opposite one.

This is a real language problem and if Znk answers you can take his answer to the bank.

znk

Answer:

[tex]\boxed{\text{(b) (ii) 0.34 V}}[/tex]

Explanation:

If electrons flow from Pb to X through the external circuit, the Pb electrode must be the anode.

The standard reduction potential for Pb²⁺ is -0.126 V.

When we write the half-reaction for the oxidation, the standard oxidation potential for Pb must be 0.126 V.

The cell reactions are:

                                                      E°/V  

  Anode: Pb ⇌ Pb²⁺ + 2e⁻           0.126

Cathode: X²⁺ + 2e⁻ ⟶ X                   x

Overall: Pb + X²⁺ ⟶ Pb²⁺ + X     0.47

0.126 + x = 0.47

x = 0.47 – 0.126 = 0.34 V

[tex]\text{The reaction at the X electrode is a reduction,}\\\text{so its standard reduction potential is }\boxed{\textbf{0.34 V}}[/tex]

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