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General Chemistry Principles, Patterns, and Applications, 2011

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We now balance the O atoms by adding H 2O—in this case, to the right side of the reduction half-reaction.<br />

Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step.<br />

reduction: Cr 2 O 7 2− (aq) → 2Cr 3+ (aq) + 7H 2 O(l)<br />

Next we balance the H atoms by adding H + to the left side of the reduction half-reaction. Again, we can ignore<br />

the oxidation half-reaction.<br />

reduction: Cr 2 O 7 2− (aq) + 14H + (aq) → + 2Cr 3+ (aq) + 7H 2 O(l)<br />

Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr +6 to 2Cr +3 ) has a<br />

+12 charge on the left <strong>and</strong> a +6 charge on the right, so six electrons are needed to balance the charge. The<br />

oxidation half-reaction (2I − to I 2) has a −2 charge on the left side <strong>and</strong> a 0 charge on the right, so it needs two<br />

electrons to balance the charge:<br />

reduction: Cr 2 O 7 2− (aq) + 14H + (aq) + 6e − → 2Cr 3+ (aq) + 7H 2 O(l)oxidation: 2I − (aq) → I 2 (aq) + 2e −<br />

Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation halfreaction<br />

by 3:<br />

oxidation: 6I − (aq) → 3I 2 (s) + 6e −<br />

Step 5: Adding the two half-reactions <strong>and</strong> canceling substances that appear in both reactions,<br />

Cr 2 O 7 2− (aq) + 14H + (aq) + 6I − (aq) → 2Cr 3+ (aq) + 7H 2 O(l) + 3I 2 (aq)<br />

Step 6: This is the same equation we obtained using the first method. Thus the charges <strong>and</strong> atoms on each<br />

side of the equation balance.<br />

Exercise<br />

Copper is commonly found as the mineral covellite (CuS). The first step in extracting the copper is to dissolve<br />

the mineral in nitric acid (HNO 3), which oxidizes sulfide to sulfate <strong>and</strong> reduces nitric acid to NO:<br />

CuS(s) + HNO 3 (aq) → NO(g) + CuSO 4 (aq)<br />

Balance this equation using the half-reaction method.<br />

Answer: 3CuS(s) + 8HNO 3(aq) → 8NO(g) + 3CuSO 4(aq) + 4H 2O(l)<br />

Calculating St<strong>and</strong>ard Cell Potentials<br />

The st<strong>and</strong>ard cell potential for a redox reaction (E°cell) is a measure of the tendency of reactants in their<br />

st<strong>and</strong>ard states to form products in their st<strong>and</strong>ard states; consequently, it is a measure of the driving force<br />

for the reaction, which earlier we called voltage. We can use the two st<strong>and</strong>ard electrode potentials we<br />

found earlier to calculate the st<strong>and</strong>ard potential for the Zn/Cu cell represented by the following cell<br />

diagram:<br />

Saylor URL: http://www.saylor.org/books<br />

Saylor.org<br />

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