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Gwar
2 months ago
15

Given two half reactions as follows: A2+ → 2 A2+ + 3 e− 4 e− + B → B4− What would you multiply each half-reaction by, to cancel

out the electrons?
Chemistry
1 answer:
Tems11 [2.7K]2 months ago
5 0
To achieve the cancellation of electrons, the oxidation half-reaction needs to be multiplied by 4 while the reduction half-reaction must be multiplied by 3. Explanation: The oxidation reaction accounts for the loss of electrons, increasing the oxidation state, while the reduction implies gaining electrons, leading to a decrease in oxidation state. The respective half-reactions illustrate this, confirming that multiplying the oxidation by 4 and the reduction by 3 achieves the desired effect.
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Details:

The equation to calculate work done is defined as follows.

W = -k \frac{q_{1}q_{2}}{d}

where, k = proportionality constant = 8.99 \times 10^{9} Jm/C^{2}

q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C

q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C

d = separation distance = 0.45 nm = 0.45 \times 10^{-9} m

Now we will insert the given values into the formula above to compute the work done as follows.

W = -k \frac{q_{1}q_{2}}{d}

= \frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}

= 3.68 \times 10^{-18} J

Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is 3.68 \times 10^{-18} J.

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2 months ago
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