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    Redox Reactions and Voltaic Cell

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    Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.)

    (a) NO2-(aq) + Cr2O72-(aq) → Cr3+ (aq) + NO3-(aq) (acidic solution)
    (b) S(s) + HNO3(aq) → H2SO3(aq) + N2O(g) (acidic solution)
    (c) Cr2O72-(aq) + CH3OH(aq) → HCO2H(aq) + Cr3+ (aq) (acidic solution)
    (d) BrO3-(aq) + N2H4(g) → Br-(aq) + N2(g) (acidic solution)
    (e) NO2-(aq) + Al(s) → NH4+(aq) + AlO2-(aq) (basic solution)
    (f) H2O2(aq) + ClO2(aq) → ClO2-(aq) + O2(g) (basic solution)

    A voltaic cell similar to that shown in Figure 20.5 is constructed. One half-cell consists of an aluminum strip placed in a solution of Al(NO3)3, and the other has a nickel strip placed in a solution of NiSO4. The overall cell reaction is
    2 Al(s) + 3 Ni2+(aq) → 2 Al3+(aq) + 3 Ni(s)

    (a) What is being oxidized, and what is being reduced?
    (b) Write the half-reactions that occur in the two half-cells.
    (c) Which electrode is the anode, and which is the cathode?
    (d) Indicate the signs of the electrodes.
    (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum?
    (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

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    Solution Preview

    Chem question
    Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.)
    (a) NO2-(aq) + Cr2O72-(aq) → Cr3+ (aq) + NO3-(aq) (acidic solution)
    3NO2- + Cr2O72- + 8H+ = 3NO3- + 2Cr3+ + 4H2O
    N is going from +3 to + 5 state, so it is getting oxidized, so NO2- is reducing agent
    Cr is going from +6 to +3 state, so it is getting reduced, so Cr2O72- is oxidizing agent

    (b) S(s) + HNO3(aq) → H2SO3(aq) + N2O(g) (acidic solution)
    S(s) + 2HNO3(aq) + 4H+ = H2SO3(aq) + N2O(g) + 2H2O
    S is going from 0 to +4 state, so it is getting oxidized, so S is reducing agent
    N is going from +5 to +1 state, so it is getting reduced, so HNO3 is oxidizing agent

    (c) Cr2O72-(aq) + CH3OH(aq) → HCO2H(aq) + ...

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