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26 September, 12:01

A galvanic cell is powered by the following redox reaction:

3Br2 (l) + 2MnO2 (s) + 8OH^ (-) (aq) = 6Br^ (-) (aq) + 2MnO4^ (-) (aq) + 4H2O (l)

a. Write a balanced equation for the half-reaction that takes place at the cathode.

b. Write a balanced equation for the half-reaction that takes place at the anode.

c. Calculate the cell voltage under standard conditions.

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  1. 26 September, 12:17
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    3Br2 (l) + 2MnO2 (s) + 8OH^ (-) (aq) - -> 6Br^ (-) (aq) + 2MnO4^ (-) (aq) + 4H2O (l)

    The cathode is where reduction takes place and oxidation takes place at the anode.

    a. Write a balanced equation for the half-reaction that takes place at the cathode.

    This means we have to write out the reduction reaction here.

    The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation; a decrease in oxidation state is known as a reduction.

    3Br2 (l) - -> 6Br^ (-) (aq)

    Reduction is the gain of electrons so we have;

    3Br2 (l) + 6e^ (-) - -> 6Br^ (-) (aq)

    Br2 (l) + 2e^ (-) - -> 2Br^ (-) (aq)

    b. Write a balanced equation for the half-reaction that takes place at the anode.

    2MnO2 (s) + 8OH^ (-) (aq) - -> 2MnO4^ (-) (aq) + 4H2O (l)

    MnO2 (s) + 4OH^ (-) (aq) - -> MnO4^ (-) (aq) + 2H2O (l) + 3e^ (-)

    c. Calculate the cell voltage under standard conditions.

    We do this by Looking up the standard potential for the half-reactions.

    Reduction - -> + 1.07

    Oxidation - -> - 1.70

    Add the cell potentials to get the overall standard cell potential.

    E⁰cell = E⁰ox + E⁰red = - 1.70 + 1.07 = - 0.63 V
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