Q1. For the balanced chemical equation 2KMnO4 + 10Nal + 8 H2S04 512 + 2 MnSO4 +

Question

Q1. For the balanced chemical equation 2KMnO4 + 10Nal + 8 H2S04 512 + 2 MnSO4 + 5 Na2SO4 + K2SO4 + 8 H2O What are the oxidizing agents? What are the reducing agents? Which atom is being oxidized? Which is being reduced? 02. Complete the chart Voltaic cell Electrolytic cell oxidation occurs at: reduction occurs at: value of AG: charge on anode: charge on cathode: Q3. Given the cell notation: Pt | Fet2(aq), Fet3+(aq) | Br20) Br(aq) |Pt What are the two half reactions taking place? Which is the anode, which is the cathode? Draw the Voltaic Cell.

Explanation / Answer

Reduction = species that GAINS electrons

Oxidation = process in which a specie will LOSS electrons

Reducing agent = The species that favors reduction, i.e. it will oxidize in order to reduce another species

Oxidizing agent = The species that favors oxidation, i.e. it will reduce in order to oxidise another species

then..

Na = remains with +1

H remains with +1

SO4 = remains as -2, then S and O are not present in redox

I = goes from -1 in NaI, to 0 in I2 --> it is being oxidized, it is the reducing agent

MnO4- = Mn+7

Mn goes to +2 in MnSO4 --> it is gaining 5 e- then it is being reduced, it is the oxidizing agent

summary:

oxidizing agent --> KMnO4

reducing agent --> NaI

atom oxidized = I

atom reduced = Mn

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