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Alborosie
3 years ago
7

7.457kCal, convert to KJ [18] = 4.184kCan]

Chemistry
1 answer:
777dan777 [17]3 years ago
6 0

Answer:

can i see a image please

Explanation:

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Answer:

\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O.

Explanation:

Consider the oxidation state on each of the element:

Left-hand side:

  • O: -2 (as in most compounds);
  • Cr: \displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6;
  • Fe: +2 (from the charge of the ion);

Right-hand side:

  • Cr: +3;
  • Fe: +3.

Change in oxidation state:

  • Each Cr atom: decreases by 3 (reduction).
  • Each Fe atom: increases by 1 (oxidation).

Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.

Assume that the coefficient of the most complex species \rm Cr_2O_7^{2-} is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.

Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain

  • two Cr atoms,
  • six Fe atoms, and
  • seven O atoms.

O atoms seldom appear among the products in acidic environments; they rapidly combine with \rm H^{+} ions to produce water \rm H_2O. Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen \rm H^{+} ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.

\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O.

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