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A voltaic cell is constructed with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell. Measurement shows that the silver electrode is positive.
(a) Write balanced half-reactions and the overall spontaneous reaction. (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients.)
reduction:
__Ag+(aq) + __e- -> ___(s)
oxidation:
__(s) ->__Pb2+(aq) +__e-
overall reaction:
__Ag+(aq) +__(s) ->__(s) + __ Pb2+(aq)

(b) The cation flow is towards which electrode?
lead OR silver
What process occurs at the lead electrode?
oxidation OR reduction
In which direction do the electrons flow?
from the lead electrode to the silver electrode OR from the silver electrode to the lead electrode

Respuesta :

IthaloAbreu

Answer:

a) Reduction:

Ag⁺(aq) + e⁻ → Ag(s)

Oxidation:

Pb(s) → Pb⁺²(aq) + 2e⁻

Overall reaction:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺

b) Silver; oxidation;  from the lead electrode to the silver electrode.

Explanation:

a) Ag⁺ had lost 1 electron, so need to gain 1 electron to become Ag(s). Pb needs to lose 2 electrons to become Pb⁺².

Reduction:

Ag⁺(aq) + e⁻ → Ag(s)

Oxidation:

Pb(s) → Pb⁺²(aq) + 2e⁻

Overall reaction:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺ (it will need 2Ag⁺ to gaind the 2 electrons released by Pb)

b) The cation formed in the redox reaction is Pb²⁺, so, to equilibrate the charges, it will flow towards the silver (Ag) electrode.

The lead (Pb) is being oxidized, so oxidation is happening at it.

The electrons flow from the oxidation (anode) to the reduction (cathode), so they flow from the lead electrode to the silver electrode.

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