Question

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) + 2NO_3^-(aq) + 8H^+(aq) --> 3Sn_2+(aq) + 2NO(g) + 4H_2O(I).

1. Which element in the cell undergoes oxidation?
- Sn
- N
- O
- H

2. Which electrode is the anode?
- tin
- platinum

Answer 1

1. Sn 2. Tin

Step-by-step explanation:

Observe the reaction,

oxidation number of Sn is zero in reaction part and in the product part the oxidation number of Sn is +2. So we see that the oxidation number rises and oxidation number rises only when oxidation happens.

Accirding to the series of reactivity, tin oxidizes faster than Platinum so tin will be placed in anode as oxidation happens in anode.

Answer 2

1.Sn

2.Sn

Step-by-step explanation:

1. Sn oxidizes to Sn2+

The oxidation number increases from 0 to +2 by removing 2 electrons

Sn ----> Sn²+ & 2e

(Oxidation is the removal of electrons)

2. In a galvanic cell, oxidation takes place at the anode, hence tin is the anode