Consider a galvanic cell with a beaker of sulfuric acid and a beaker of nitric acid. The sulfuric acid beaker contains a strip of tin, and the nitric acid cell contains a strip of platinum. A wire runs between the strips. The reaction that occurs is as follows:3Sn(s) + 2NO3–(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)I understand the concept of the galvanic cell, but I don't understand the reactions or how to figure out half reactions.

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04-11-2022
Chemistry

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Consider a galvanic cell with a beaker of sulfuric acid and a beaker of nitric acid. The sulfuric acid beaker contains a strip of tin, and the nitric acid cell contains a strip of platinum. A wire runs between the strips. The reaction that occurs is as follows:3Sn(s) + 2NO3–(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)I understand the concept of the galvanic cell, but I don't understand the reactions or how to figure out half reactions.

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NorrinN748908 NorrinN748908 · Answer 1 · 2022-11-04T04:17:11-04:00

First of all, we have a REDOX reaction here because we have loss and gain of electrons. The process of electrolysis is accompanied by the redox reaction.

Explanation:

3Sn(s) + 2NO3–(aq) + 8H+(aq) → 3Sn2+(aq) + 2NO(g) + 4H2O(l)

On the left, we have Sn (s) with an oxidation number equal to 0, then on the right we have Sn+2, so it lost electrons because it gained a positive charge. If we lose electrons, then we have oxidation, and it takes place at the anode. Thus, the anode is composed of tin.

Reaction at the anode:

Sn (s) => Sn+2 (aq) + 2e-

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We have on the left N in NO3 which oxidation number is +6, and on the left N in NO has +2 as oxidation number.

We can say that the oxidation number decreases from +6 to +2. Therefore, we have a reduction of NO3 and it takes place at the cathode. The cathode is composed of Pt.

Reaction at the cathode:

NO3- (aq) + 4H+ (aq) + 3e- => NO (g) + 2 H2O (l)

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To get to the complete reaction, we must multiply by 3 the reaction at the anode, and by 2 the reaction at the cathode. After that, we add both reactions together.