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Consider the following half-cell reactions.Al(s) → Al3+(aq) + 3e-Cu2+(aq) + 2e...


Question

Consider the following half-cell reactions.

Al(s) → Al3+(aq) + 3e-

Cu2+(aq) + 2e- → Cu(s)

The overall equation for the reaction is

Options

A) Al(s) + Cu2+(aq) → Al3+(aq) + Cu(s)

B) 2Al(s) + Cu2+(aq) → 2Al(aq) + Cu(s)

C) 2Al(s) + 3Cu2+(aq) → 3Cu(s) + 2Al3+(aq)

D) 3Al(s) + 2Cu2+(aq) → Cu(s) + 3Al3+(aq)


The correct answer is C.

Explanation:

Al(s) → Al3+(aq) + 3e- -------(i)
Cu2+(aq) + 2e- → Cu(s) -------(ii)
To balance the electrons, multiply equation (i) by 2 and equation (ii) by 3 and then add the resulting equations
2Al(s) → 2Al3+(aq) + 3e-
3Cu2+(aq) + 6e- → 3Cu(s)
2Al(s) + 3Cu2+(aq) → 3Cu(s) + 2Al3+(aq)


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Discussion (2)

  • Praise

    It Should be this,since you've multiplied equation (i) by 2
    6e– for the first equation

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  • Al(s) → Al3+(aq) + 3e- -------(i)
    Cu2+(aq) + 2e- → Cu(s) -------(ii)
    To balance the electrons, multiply equation (i) by 2 and equation (ii) by 3 and then add the resulting equations
    2Al(s) → 2Al3+(aq) + 3e-
    3Cu2+(aq) + 6e- → 3Cu(s)
    2Al(s) + 3Cu2+(aq) → 3Cu(s) + 2Al3+(aq)

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