Section Test:Electrochemical cells

From MyMCAT

Jump to: navigation, search

1. In a lead-acid battery, the following reaction occurs during discharge, Pb + PbO2 + H2SO4 --> 2PbSO4 + 2H2O. Which of the following species is undergoing oxidation?

Pb
PbO2
H2O
SO42-
→ In assigning oxidation numbers, SO4 gets -2, thus each H gets +1, this does not change when H2O is formed as the O is -2 and the Hs are still +1. The Pb however is 0 and becomes +2 to balance the SO4, thus the Pb is oxidized.

2. What type of reaction is occurring when old iron gates rust due to moisture and oxygen,

neutralization
reduction-oxidation
saponification
titration
→ Rusting is an example of corrosion, which is caused by air combining with the metal to form a metal oxide. These reactions occur because the metal loses electrons to the oxygen. (It went from an oxidation number of zero to some + number to balance the oxyden's -2s.)

3. Given the redox reaction, 2I-(aq) + Br2(l) -> 2Br-(aq) + I2(s), what occurs during this reaction?

The I- ion is oxidized, and its oxidation number increases.
The I- ion is oxidized, and its oxidation number decreases.
The I- ion is reduced, and its oxidation number increases.
The I- ion is reduced, and its oxidation number decreases.
→ I- has an oxidation number of -1. After the reaction, I2(s) has an oxidation number of zero, thus each I has an oxidation number of zero. This change from -1 to 0 is an increase in oxidation number and thus I undergoes oxidation.

4. Which reaction correctly represents a reduction half-cell?

Cr(s) -> Cr3+ + 3e-
Cr3++ -> Cr(s) + 3e-
Cr3+ + 3e- -> Cr(s)
Cr(s) + 3e- -> Cr3+

5. Which species undergoes oxidation in the reaction, H2 + Sn4+ -> 2H+ + Sn2+?

H2
Sn4+
2H+
Sn2+

6. What is the oxidation number of P in H4P2O7?

+1
+3
+4
+5
→ Overall, we know that the molecule has an oxidation state of 0 (because it is neutral charge).
→ Each H will contribute +1 to this and each O will contribute -2 (these are the standard values for H and O), therefore we would expect that all the H's would give +4 and all the O's would give -14.
→ Thus we have, (+4)+2X+(-14)=0, where we need to solve for X, the unknown oxidation value for P.
→ Solving this will yield 2X = -10, or X=+5.

7. Which species is an oxidizing agent in the reaction, 2Fe2+ + Sn4+ -> 2Fe3+ + Sn2+?

Fe2+
Sn4+
Fe3+
Sn2+
→ An oxidizing agent causes another species to oxidize, but it itself will undergo reduction. For reduction, we are looking for a gain in electrons, or a change in the charge of a species towards the more negative. Sn here goes from +4 to +2, which implies it has gained two electrons (to become more negative) thus Sn4+ is the reducing agent at play here. One could further verify this, but checking that the other element, Fe2+, is oxidized.

Your score is 0 / 0
Retrieved from "http://www.mymcat.com/wiki/Section_Test:Electrochemical_cells"