M(C2H2O)= 12.0*2 +1.0*2 +16.0 = 42 g/mol is a molar mass for empirical formula.
120.6g/mol/42g/mol ≈ 3
So, empirical formula should be increased 3 times,
and molecular formula is C6H6O3.
Answer is D.
Answer:
2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺
Explanation:
To balance a redox reaction in an acidic medium, we simply follow some rules:
- Split the reaction into an oxidation and reduction half.
- By inspecting, balance the half equations with respect to the charges and atoms.
- In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
- Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
- Multiply both equations with appropriate factors to balance the electrons in the two half equations.
- Add up the balanced half equations and cancel out any specie that occur on both sides.
- Check to see if the charge and atoms are balanced.
Solution
Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺
The half equations:
Zn → Zn²⁺ Oxidation half
MnO₄⁻ → Mn²⁺ Reduction half
Balancing of atoms(in acidic medium)
Zn → Zn²⁺
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balancing of charge
Zn → Zn²⁺ + 2e⁻
MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O
Balancing of electrons
Multiply the oxidation half by 5 and reduction half by 2:
5Zn → 5Zn²⁺ + 10e⁻
2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O
Adding up the two equations gives:
5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O
The net equation gives:
5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O
Cofactors:
A. Coenzyme A (CoA-SH)
B. NAD+
C. Thiamine pyrophosphate (TPP)
D. FAD
E. Lipoic acid in oxidized form
Roles:
E... Attacks and attaches to the central carbon in pyruvate.
A...Oxidizes FADH2.
C...Accepts the acetyl group from reduced lipoic acid.
D... Oxidizes the reduced form of lipoic acid.
B... Initial electron acceptor in oxidation of pyruvate.
Temperature of the environment, surrounding atoms and size of atom are all factors which affect removal of electron. Increase in temperature increases the probability of removal of electron. But size of the atom plays very important role in removal of electrons. In case of large sized atom, the nuclear attraction of nucleus on valence electrons decreases which eases the removal of electrons. Hope this helps.
