
(a)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can gain or lose the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.
(b)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.
(c)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can be gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state
(d)
Interpretation: The oxidation state of given atom in the given molecule has to be calculated.
Concept introduction: In general the atom can be gain or loss the electron we get corresponding negative or positive ions that positive or negative charge is also called oxidation state.

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Chapter 21 Solutions
Student Solutions Manual for Ebbing/Gammon's General Chemistry, 11th
- a. Explain Why electron withdrawing groups tend to be meta-Directors. Your answer Should lyclude all apropriate. Resonance contributing Structures fo. Explain why -ll is an outho -tura drccton even though chlorine has a very High Electronegativityarrow_forward9. Write Me product as well as the reaction Mechanism For each of the Following Vanctions +H₂504 4.50+ T C. +212 Fellz 237 b. Praw the potential energy Diagrams For each OF Mese Rauctions and account For any differences that appear in the two potential Puergy Diagrams which of here two reactions 19 Found to be Reversable, Rationalice your answer based upon the venation mechanisms and the potential energy diagrams.arrow_forward9. Write Me product as well as the reaction Mechanism For each of the Following Veritious +H2504 4.50+ + 1/₂ Felly ◎+ 7 b. Praw he potential energy Diagrams For each OF Mese Ronctions and account for any differences that appeak in the two potential Puergy Diagramsarrow_forward
- Draw the major product of this reaction. Ignore inorganic byproducts. Incorrect, 3 attempts remaining 1. excess Br2, NaOH 2. neutralizing workup Qarrow_forwardGiven the electrode Pt | Ag | Ag+ (aq), describe it.arrow_forwardAt 25°C, the reaction Zn2+ + 2e ⇄ Zn has a normal equilibrium potential versus the saturated calomel electrode of -1.0048 V. Determine the normal equilibrium potential of Zn versus the hydrogen electrode.Data: The calomel electrode potential is E° = 0.2420 V versus the normal hydrogen electrode.arrow_forward
- Electrochemistry. State the difference between E and E0.arrow_forwardIn an electrolytic cell, the positive pole is always assumed to be on the right side of the battery notation. Is that correct?arrow_forwardIn an electrolytic cell, the positive pole is always assumed to be on the right side of the battery. Is that correct?arrow_forward
- Calculate the free energy of formation of 1 mol of Cu in cells where the electrolyte is 1 mol dm-3 Cu2+ in sulfate solution, pH 0. E° for the Cu2+/Cu pair in this medium is +142 mV versus ENH.Assume the anodic reaction is oxygen evolution.Data: EH2 = -0.059 pH (V) and EO2 = 1.230 - 0.059 pH (V); 2.3RT/F = 0.059 Varrow_forwardIf the normal potential for the Fe(III)/Fe(II) pair in acid at zero pH is 524 mV Hg/Hg2Cl2 . The potential of the saturated calomel reference electrode is +246 mV versus the NHE. Calculate E0 vs NHE.arrow_forwardGiven the galvanic cell whose scheme is: (-) Zn/Zn2+ ⋮⋮ Ag+/Ag (+). If we know the normal potentials E°(Zn2+/Zn) = -0.76V and E°(Ag+/Ag) = 0.799 V. Indicate the electrodes that are the anode and the cathode and calculate the E0battery.arrow_forward
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