The diagram shows a portable electrolysis unit designed for a science centre display. It uses a battery to pass current through water containing a few drops...

Assessment: Chemistry 9202 | Paper 1 Mock 01 | Written Paper 1 Subject: Chemistry - 9202

Question 1 Report

The diagram shows a portable electrolysis unit designed for a science centre display. It uses a battery to pass current through water containing a few drops of dilute sulfuric acid. The acid provides ions so that the solution conducts electricity. Gas is collected in two inverted measuring tubes. Hydrogen is made at the negative electrode and oxygen is made at the positive electrode. Table 1 gives results from one 15-minute display. The electrical energy supplied by the battery is also shown. The gases are allowed to cool to room temperature before their volumes are read.

Table 1
GasVolume collected / cm3
Hydrogen12
Oxygen6
Electrical energy supplied1800 J

(a) Name the positive electrode in this electrolysis cell. [1]
(b) Use Table 1 to give the simplest whole-number ratio of hydrogen volume to oxygen volume. [2]
(c) Describe the overall energy transfer in the electrolysis of water. [2]
(d) Explain why hydrogen forms at the negative electrode. [3]
(e) Use Table 1 to calculate the electrical energy supplied for each cm3 of gas collected in total. [3]

Answer Details

Electrolysis of water: Electrolysis uses electrical energy to decompose water. Positive hydrogen ions move to the negative electrode, where they gain electrons.

  1. (a) The positive electrode is the anode. [1]
  2. (b) \[12:6=2:1\] The simplest hydrogen:oxygen volume ratio is 2 : 1. [2]
  3. (c) Electrical energy from the battery is transferred and stored as chemical energy in hydrogen and oxygen, or used to decompose water. [2]
  4. (d) Hydrogen ions, ? ? H+ ions, move to the negative electrode. They gain electrons, and hydrogen atoms then join in pairs to form H2 molecules. [3]
  5. (e) Total gas volume: \[12+6=18\ \mathrm{cm^3}\] Energy per cubic centimetre: \[\frac{1800\ \mathrm{J}}{18\ \mathrm{cm^3}}=100\ \mathrm{J\ cm^{-3}}\] [3]

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